JEE Advanced Maths PYQ – Conic Sections (Ellipse) | Previous Year Questions with Solutions

Q 1 :

Let $S$ denote the locus of the point of intersection of the pair of lines $4 x - 3 y = 12 α, 4 α x + 3 α y = 12$ , where $α$ varies over the set of non-zero real numbers. Let $T$ be the tangent to $S$ passing through the points $(p, 0)$ and $(0, q), q > 0$ and parallel to the line $4 x - \frac{3}{\sqrt{2}} y = 0$ . Then the value of $p q$ is [2025]

$- 6 \sqrt{2}$
$- 3 \sqrt{2}$
$- 9 \sqrt{2}$
$- 12 \sqrt{2}$

1

2

3

4

(1)

$4 x - 3 y = 12 α . . . (i)$

$4 x + 3 y = \frac{12}{α} . . . (i i)$

$For elimination of α, multiply equations (i) and (ii)$

$\Rightarrow 16 x^{2} - 9 y^{2} = 144$ $\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 (Hyperbola)$

$Since, T is parallel to 4 x - \frac{3 y}{\sqrt{2}} = 0$

$\Rightarrow m = \frac{4}{3 / \sqrt{2}} = \frac{4 \sqrt{2}}{3}$

$So equation of tangent:$

$y = m x + \sqrt{a^{2} m^{2} - b^{2}} as q > 0$

$\Rightarrow y = \frac{4 \sqrt{2}}{3} x + \sqrt{9 \times \frac{16 \times 2}{9} - 16}$

$\Rightarrow y = \frac{4 \sqrt{2} x}{3} + 4$ $\Rightarrow (p, 0) \Rightarrow 0 = \frac{4 \sqrt{2} p}{3} + 4 \Rightarrow p = - \frac{3}{\sqrt{2}}$

$and (0, q) \Rightarrow q = 4;$ $∴$ $p q = \frac{- 3}{\sqrt{2}} \times 4 = - 6 \sqrt{2}$

Q 2 :

Consider the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ . Let $S (p, q)$ be a point in the first quadrant such that $\frac{p^{2}}{9} + \frac{q^{2}}{4} > 1$ . Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$ -coordinate and $O$ be the center of the ellipse. If the area of the triangle $∆ O R T$ is $\frac{3}{2}$ , then which of the following options is correct? [2024]

$q = 2, p = 3 \sqrt{3}$
$q = 2, p = 4 \sqrt{3}$
$q = 1, p = 5 \sqrt{3}$
$q = 1, p = 6 \sqrt{3}$

1

2

3

4

(1)

$Given that \frac{p^{2}}{9} + \frac{q^{2}}{4} > 1 then S (p, q) lies outside the ellipse$

$SA is tangent \Rightarrow q = 2$

$Area of ∆ O R T = \frac{3}{2}$

$\Rightarrow$ $| \frac{1}{2} \times O R \times Q T |$ $= \frac{3}{2} \Rightarrow$ $| \frac{1}{2} \times 3 \times β |$ $= \frac{3}{2} \Rightarrow β = - 1$

$∴$ $\frac{α^{2}}{9} + \frac{β^{2}}{4} = 1 \Rightarrow \frac{α^{2}}{9} = 1 - \frac{1}{4} = \frac{3}{4}$

$\Rightarrow α^{2} = \frac{27}{4} \Rightarrow α = \frac{3 \sqrt{3}}{2}$

$Tangent at T$

$T = 0$

$\frac{x \cdot \frac{3 \sqrt{3}}{2}}{9} + \frac{y (- 1)}{4} = 1 |_{(p, 2)}$

$\Rightarrow \frac{p \sqrt{3}}{6} - \frac{1}{2} = 1 \Rightarrow \frac{p \sqrt{3}}{6} = \frac{3}{2} \Rightarrow p = 3 \sqrt{3}$

$∴$ $p = 3 \sqrt{3}, q = 2$

Q 3 :

The ellipse $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point (0, 4) circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is [2012]

$\frac{\sqrt{2}}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{3}{4}$

1

2

3

4

(3)

$As rectangle A B C D circumscribes the ellipse \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1,$

$∴$ $A = (3, 2)$

$Let the ellipse circumscribing the rectangle A B C D is$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (i)$

$Given that ellipse (i) passes through (a, 4) \Rightarrow b^{2} = 16$

$Also ellipse (i) passes through A (3, 2)$

$∴$ $\frac{9}{a^{2}} + \frac{4}{16} = 1 \Rightarrow \frac{9}{a^{2}} = 1 - \frac{1}{4} = \frac{3}{4}$

$\Rightarrow a^{2} = 12$

$∴$ $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{12}{16}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

Q 4 :

The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $x$ -axis at $Q$ . If $M$ is the midpoint of the line segment $P Q$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points [2009]

$(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7})$
$(\pm \frac{3 \sqrt{5}}{2}, \pm \sqrt{\frac{19}{4}})$
$(\pm 2 \sqrt{3}, \pm \frac{1}{7})$
$(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7})$

1

2

3

4

(3)

$Given ellipse is \frac{x^{2}}{4^{2}} + \frac{y^{2}}{2^{2}} = 1$

$∴$ $a^{2} = 16, b^{2} = 4 \Rightarrow e^{2} = 1 - \frac{4}{16} = \frac{3}{4} \Rightarrow e = \frac{\sqrt{3}}{2}$

$Let P (4 \cos θ, 2 \sin θ) be any point on the ellipse, then equation of normal at P is$

$4 x \sin θ - 2 y \cos θ = 12 \sin θ \cos θ$

$\Rightarrow \frac{x}{3 \cos θ} - \frac{y}{6 \sin θ} = 1$

$∴$ $Q, the point where normal at P meets x-axis, has coordinates (3 \cos θ, 0)$

$∴$ $Mid point of P Q is M (\frac{7 \cos θ}{2}, \sin θ)$

$For locus of point M we consider$

$x = \frac{7 \cos θ}{2}, y = \sin θ \Rightarrow \cos θ = \frac{2 x}{7}, \sin θ = y$

$Since \sin^{2} θ + \cos^{2} θ = 1$

$∴$ $\frac{4 x^{2}}{49} + y^{2} = 1 . . . (i)$

$Also the latus rectum of given ellipse is$

$x = \pm a e = \pm 4 \times \frac{\sqrt{3}}{2} \Rightarrow x = \pm 2 \sqrt{3} . . . (i i)$

$On solving equations (i) and (ii), we get$

$\frac{4 \times 12}{49} + y^{2} = 1 \Rightarrow y^{2} = \frac{1}{49} or y = \pm \frac{1}{7}$

$∴$ $The required points are (\pm 2 \sqrt{3}, \pm \frac{1}{7}) .$

Q 5 :

The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^{2} + 9 y^{2} = 9$ meets its auxiliary circle at the point $M$ . Then the area of the triangle with vertices at $A, M$ and the origin $O$ is [2009]

$\frac{31}{10}$
$\frac{29}{10}$
$\frac{21}{10}$
$\frac{27}{10}$

1

2

3

4

(4)

$Given ellipse is x^{2} + 9 y^{2} = 9 \Rightarrow \frac{x^{2}}{3^{2}} + \frac{y^{2}}{1^{2}} = 1$

$∴$ $Coordinates of A and B are (3, 0) and (0, 1) respectively$

$∴$ $Equation of A B is \frac{x}{3} + \frac{y}{1} = 1$

$\Rightarrow x + 3 y - 3 = 0 . . . (i)$

$and equation of auxiliary circle of given ellipse is$

$x^{2} + y^{2} = 9 . . . (i i)$

$On solving equations (i) and (ii), we get the point M where line A B meets the auxiliary circle.$

$Putting x = 3 - 3 y from (i) in (ii), we get$

${(3 - 3 y)}^{2} + y^{2} = 9$

$\Rightarrow 9 - 18 y + 9 y^{2} + y^{2} = 9$ $\Rightarrow 10 y^{2} - 18 y = 0$

$\Rightarrow y = 0, \frac{9}{5} \Rightarrow x = 3, \frac{- 12}{5}$

$Clearly M (\frac{- 12}{5}, \frac{9}{5})$

$∴$ $Area of ∆ O A M = \frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ 3 & 0 & 1 \\ \frac{- 12}{5} & \frac{9}{5} & 1 \end{matrix} |$ $= \frac{27}{10}$

Q 6 :

The minimum area of triangle formed by the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and coordinate axes is [2005]

$a b sq. units$
$\frac{a^{2} + b^{2}}{2} sq. units$
$\frac{{(a + b)}^{2}}{2} sq. units$
$\frac{a^{2} + a b + b^{2}}{3} sq. units$

1

2

3

4

(1)

$Any tangent to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at P (a \cos θ, b \sin θ) is$ $\frac{x \cos θ}{a} + \frac{y \sin θ}{b} = 1$

$It meets coordinate axes at A (a \sec θ, 0) and B (0, b c o s e c θ)$

$∴$ $Area of ∆ O A B = \frac{1}{2} \times a \sec θ \times b c o s e c θ = \frac{a b}{\sin 2 θ}$

$For area of ∆ O A B to be minimum, \sin 2 θ should be maximum, i.e., 1$

$∴$ $Minimum area of ∆ O A B = a b sq. units.$

Q 7 :

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is [2004]

$\frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1$
$\frac{1}{4 x^{2}} + \frac{1}{2 y^{2}} = 1$
$\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$
$\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$

1

2

3

4

(1)

$Any tangent to ellipse \frac{x^{2}}{2} + \frac{y^{2}}{1} = 1 is$

$\frac{x \cos θ}{\sqrt{2}} + y \sin θ = 1$ $which makes intercept A B between the coordinate axes.$

$∴$ $A (\sqrt{2} \sec θ, 0), B (0, \cos e c θ)$

$If (h, k) be the mid-point of A B, then$

$2 h = \sqrt{2} \sec θ, 2 k = c o s e c θ$

$\Rightarrow \cos θ = \frac{1}{\sqrt{2} h}, \sin θ = \frac{1}{2 k}$

$Now \cos^{2} θ + \sin^{2} θ = 1$

$\Rightarrow {(\frac{1}{\sqrt{2} h})}^{2} + {(\frac{1}{2 k})}^{2} = 1$ $\Rightarrow \frac{1}{2 h^{2}} + \frac{1}{4 k^{2}} = 1$

$∴$ $Required locus is \frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1$

Q 8 :

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ , is [2003]

$\frac{27}{4} sq. units$
$9 sq. units$
$\frac{27}{2} sq. units$
$27 sq. units$

1

2

3

4

(4)

$Given equation of ellipse is \frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$

$∴$ $a^{2} = 9, b^{2} = 5 \Rightarrow e = \sqrt{1 - \frac{5}{9}} = \frac{2}{3}$

$∴$ $End point of latus rectum in first quadrant is L (2, \frac{5}{3})$

$Equation of tangent at L is \frac{2 x}{9} + \frac{y}{3} = 1,$ $which meets x-axis at A (\frac{9}{2}, 0) and y-axis at B (0, 3)$

$∴$ $Area of ∆ O A B = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$

$Now by symmetry area of quadrilateral A B C D$

$= 4 \times Area (∆ O A B) = 4 \times \frac{27}{4} = 27 sq. units.$

Q 9 :

Suppose that the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ are $(f_{1}, 0)$ and $(f_{2}, 0)$ where $f_{1} > 0$ and $f_{2} < 0$ . Let $P_{1}$ and $P_{2}$ be two parabolas with a common vertex at (0, 0) and with foci at $(f_{1}, 0)$ and $(2 f_{2}, 0)$ , respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2 f_{2}, 0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1}, 0)$ . If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$ , then the value of $(\frac{1}{m_{1}^{2}} + m_{2}^{2})$ is [2015]

(4)

$Given: Ellipse is \frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$

$\Rightarrow a = 3, b = \sqrt{5}, e = \frac{2}{3} \Rightarrow f_{1} = 2, f_{2} = - 2$

$P_{1} : y^{2} = 8 x and P_{2} : y^{2} = - 16 x$

$T_{1} : y = m_{1} x + \frac{2}{m_{1}}$

$It passes through (- 4, 0),$ $∴$ $0 = - 4 m_{1} + \frac{2}{m_{1}} \Rightarrow m_{1}^{2} = \frac{1}{2}$

$T_{2} : y = m_{2} x - \frac{4}{m_{2}}, It passes through (2, 0)$

$∴$ $0 = 2 m_{2} - \frac{4}{m_{2}} \Rightarrow m_{2}^{2} = 2$

$∴$ $\frac{1}{m_{1}^{2}} + m_{2}^{2} = 2 + 2 = 4$

Q 10 :

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If $Δ (h)$ is the area of the triangle PQR, $Δ_{1} = \max_{1 / 2 \leq h \leq 1} Δ (h) and Δ_{2} = \min_{1 / 2 \leq h \leq 1} Δ (h),$ then $\frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} =$ [2013]

(9)

$Vertical line x = h, meets the ellipse \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1 at$

$P (h, \frac{\sqrt{3}}{2} \sqrt{4 - h^{2}}) and Q (h, \frac{- \sqrt{3}}{2} \sqrt{4 - h^{2}})$

$By symmetry, tangents at P and Q will meet each other at x-axis.$

$Tangent at P is \frac{x h}{4} + \frac{y \sqrt{3}}{6} \sqrt{4 - h^{2}} = 1,$ $which meets x-axis at R (\frac{4}{h}, 0)$

$area (∆ P Q R) = \frac{1}{2} \times \sqrt{3} \sqrt{4 - h^{2}} \times (\frac{4}{h} - h)$

$Let Δ (h) = \frac{\sqrt{3} {(4 - h^{2})}^{3 / 2}}{2 h}$

$\Rightarrow \frac{d Δ}{d h} = - \sqrt{3} [\frac{\sqrt{4 - h^{2}} (h^{2} + 2)}{h^{2}}] < 0$

$∴$ $Δ (h) is a decreasing function.$

$∴$ $\frac{1}{2} \leq h \leq 1 \Rightarrow Δ_{\max} = Δ (\frac{1}{2}) and Δ_{\min} = Δ (1)$

$∴$ $Δ_{1} = Δ_{\max} = \frac{\sqrt{3}}{2} \cdot \frac{{(4 - \frac{1}{4})}^{3 / 2}}{\frac{1}{2}} = \frac{45}{8} \sqrt{5}$ $and Δ_{2} = Δ_{\min} = \frac{\sqrt{3} \cdot 3 \sqrt{3}}{2 \cdot 1} = \frac{9}{2}$

$∴$ $\frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} = 45 - 36 = 9$

Q 11 :

Let $E$ be the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 .$ For any three distinct points $P, Q$ and $Q'$ on $E$ , let $M (P, Q)$ be the midpoint of the line segment joining $P$ and $Q$ , and $M (P, Q')$ be the midpoint of the line segment joining $P$ and $Q'$ . Then the maximum possible value of the distance between $M (P, Q)$ and $M (P, Q')$ , as $P, Q$ and $Q'$ vary on $E$ , is ______. [2021]

(4)

$Let A and B be midpoints of segment P Q and P Q' respectively$

$By midpoint theorem$

$A B ∥ Q Q' and A B = \frac{1}{2} Q Q'$

$∴$ $Distance between M (P, Q) and M (P, Q') = \frac{1}{2} Q Q'$

$Since, Q, Q' must be on E, so, maximum of Q Q' = 8$

$∴$ $Maximum of A B = \frac{8}{2} = 4$

Q 12 :

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ be two distinct points on the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ such that $y_{1} > 0$ and $y_{2} > 0$ . Let $C$ denote the circle $x^{2} + y^{2} = 9,$ and $M$ be the point (3, 0). Suppose the line $x = x_{1}$ intersects $C$ at $R$ , and the line $x = x_{2}$ intersects $C$ at $S$ , such that the $y$ -coordinates of $R$ and $S$ are positive. Let $∠ R O M = \frac{π}{6}$ and $∠ S O M = \frac{π}{3},$ where $O$ denotes the origin (0, 0). Let $| X Y |$ denote the length of the line segment $X Y$ .

Then which of the following statement(s) is (are) TRUE? [2025]

The equation of the line joining P and Q is $2 x + 3 y = 3 (1 + \sqrt{3})$
The equation of the line joining P and Q is $2 x + y = 3 (1 + \sqrt{3})$
If $N_{2} = (x_{2}, 0)$ , then $3 | N_{2} Q | = 2 | N_{2} S |$
If $N_{1} = (x_{1}, 0)$ , then $9 | N_{1} P | = 4 | N_{1} R |$

1

2

3

4

Select one or more options

(1, 3)

$m_{P Q} = \frac{\sqrt{3} - 1}{\frac{3}{2} (1 - \sqrt{3})} = \frac{- 2}{3}$

$So, equation of P Q is 2 x + 3 y = 3 + 3 \sqrt{3}$

$\Rightarrow 2 x + 3 y = 3 (1 + \sqrt{3})$

$3 | N_{2} Q | = 2 | N_{2} S | \Rightarrow 3 \times \sqrt{3} = 2 \times \frac{3 \sqrt{3}}{2}$

$9 | N_{1} P | = 4 | N_{1} R | \Rightarrow 9 \times 1 = 4 \times \frac{3}{2}$

Q 13 :

Let $T_{1}$ and $T_{2}$ be two distinct common tangents to the ellipse $E : \frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ and the parabola $P : y^{2} = 12 x .$ Suppose that the tangent $T_{1}$ touches $P$ and $E$ at the points $A_{1}$ and $A_{2}$ , respectively, and the tangent $T_{2}$ touches $P$ and $E$ at the points $A_{4}$ and $A_{3}$ , respectively. Then which of the following statement(s) is (are) true? [2023]

The area of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is 35 square units
The area of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is 36 square units
The tangents $T_{1}$ and $T_{2}$ meet the $x$ -axis at the point $(- 3, 0)$
The tangents $T_{1}$ and $T_{2}$ meet the $x$ -axis at the point $(- 6, 0)$

1

2

3

4

Select one or more options

(1, 3)

$Given equation of ellipse E : \frac{x^{2}}{6} + \frac{y^{2}}{3} = 1 .$

$Tangent: y = m_{1} x \pm \sqrt{6 m_{1}^{2} + 3}$

$Equation of parabola P : y^{2} = 12 x$

$Tangent: y = m_{2} x + \frac{3}{m_{2}}$

$For common tangent$

$m_{1} = m_{2} = m (say) and \pm \sqrt{6 m_{1}^{2} + 3} = \frac{3}{m_{2}}$

$\Rightarrow m = \pm 1$

$Equation of common tangents are$

$T_{1} : y = x + 3 and T_{2} : y = - x - 3 .$

$∵$ $point of contact for parabola is (\frac{a}{m^{2}}, \frac{2 a}{m})$

$\Rightarrow A_{1} (3, 6) and A_{4} (3, - 6)$ $on solving y = x + 3 or y = - x - 3 and$

$equation of ellipse \frac{x^{2}}{6} + \frac{y^{2}}{3} = 1 we get A_{2} (- 2, 1) and A_{3} (- 2, - 1) .$

$Area of quadrilateral A_{1} A_{2} A_{3} A_{4} = \frac{1}{2} (12 + 2) \times 5$

$(∵ A_{1} A_{4} = 12, A_{2} A_{3} = 2, M N = 5)$

$= 35 sq. units.$

$Put y = 0 in T_{1} and T_{2} we get point of intersection with x-axis is (- 3, 0) .$

$Hence option (1) and (3) are correct.$

Q 14 :

Consider two straight lines, each of which is tangent to both the circle $x^{2} + y^{2} = \frac{1}{2}$ and the parabola $y^{2} = 4 x .$ Let these lines intersect at the point $Q$ . Consider the ellipse whose center is at the origin $O (0, 0)$ and whose semi-major axis is $O Q$ . If the length of the minor axis of this ellipse is $\sqrt{2}$ , then which of the following statement(s) is (are) TRUE? [2018]

For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is 1
For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$
The area of the region bounded by the ellipse between the lines $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is $\frac{1}{4 \sqrt{2}} (π - 2)$
The area of the region bounded by the ellipse between the lines $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is $\frac{1}{16} (π - 2)$

1

2

3

4

Select one or more options

(1, 3)

$Let the equation of common tangent is y = m x + \frac{1}{m}$

$∴$ $| \frac{0 + 0 + \frac{1}{m}}{\sqrt{1 + m^{2}}} |$ $= \frac{1}{\sqrt{2}}$

$\Rightarrow m^{4} + m^{2} - 2 = 0 \Rightarrow m = \pm 1$

$∴$ $Equation of common tangents are$

$y = x + 1$ and $y = - x - 1$ $∴$ $Q \equiv (- 1, 0)$

$∴$ $Equation of ellipse is \frac{x^{2}}{1} + \frac{y^{2}}{1 / 2} = 1$

$(1)$ $e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$ $and latus rectum = \frac{2 b^{2}}{a} = \frac{2 {(\frac{1}{\sqrt{2}})}^{2}}{1} = 1$

$(3)$

$∴$ $Required area = 2 \int_{1 / \sqrt{2}}^{1} \frac{1}{\sqrt{2}} \sqrt{1 - x^{2}} d x$

$= \sqrt{2} {[\frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} x]}_{1 / \sqrt{2}}^{1}$

$= \sqrt{2} [\frac{π}{4} - (\frac{1}{4} + \frac{π}{8})] = \sqrt{2} (\frac{π}{8} - \frac{1}{4}) = \frac{π - 2}{4 \sqrt{2}}$

Q 15 :

Let $E_{1}$ and $E_{2}$ be two ellipses whose centers are at the origin. The major axes of $E_{1}$ and $E_{2}$ lie along the $x$ -axis and the $y$ -axis, respectively. Let $S$ be the circle $x^{2} + {(y - 1)}^{2} = 2 .$ The straight line $x + y = 3$ touches the curves $S$ , $E_{1}$ and $E_{2}$ at $P$ , $Q$ and $R$ respectively. Suppose that $P Q = P R = \frac{2 \sqrt{2}}{3} .$ If $e_{1}$ and $e_{2}$ are the eccentricities of $E_{1}$ and $E_{2}$ , respectively, then the correct expression(s) is (are) [2015]

$e_{1}^{2} + e_{2}^{2} = \frac{43}{40}$
$e_{1} e_{2} = \frac{\sqrt{7}}{2 \sqrt{10}}$
$| e_{1}^{2} - e_{2}^{2} |$ $= \frac{5}{8}$
$e_{1} e_{2} = \frac{\sqrt{3}}{4}$

1

2

3

4

Select one or more options

(1, 2)

$Let E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a > b \dots (i)$

$and E_{2} : \frac{x^{2}}{c^{2}} + \frac{y^{2}}{d^{2}} = 1, where c < d \dots (i i)$

$Also S : x^{2} + {(y - 1)}^{2} = 2 \dots (i i i)$

$Tangent at P (x_{1}, y_{1}) to (i i i) is x + y = 3 \dots (i v)$

$On solving (i i i) and (i v), we get the point of contact P (1, 2)$

$Now, equation of tangent in parametric form,$

$\frac{x - 1}{\frac{- 1}{\sqrt{2}}} = \frac{y - 2}{\frac{1}{\sqrt{2}}} = \pm \frac{2 \sqrt{2}}{3}$ $\Rightarrow x = \frac{1}{3} or \frac{5}{3}$ and $y = \frac{8}{3} or \frac{4}{3}$

$∴$ $Q (\frac{5}{3}, \frac{4}{3}) and R (\frac{1}{3}, \frac{8}{3})$

$Now, equation of tangent to E_{1} at Q is$

$\frac{5 x}{3 a^{2}} + \frac{4 y}{3 b^{2}} = 1$ $which is identical to \frac{x}{3} + \frac{y}{3} = 1$

$\Rightarrow a^{2} = 5 and b^{2} = 4 \Rightarrow e_{1}^{2} = 1 - \frac{4}{5} = \frac{1}{5}$

$And equation of tangent to E_{2} at R is$

$\frac{x}{3 c^{2}} + \frac{8 y}{3 d^{2}} = 1, which is identical to \frac{x}{3} + \frac{y}{3} = 1$

$\Rightarrow c^{2} = 1, d^{2} = 8 \Rightarrow e_{2}^{2} = 1 - \frac{1}{8} = \frac{7}{8}$

$∴$ $e_{1}^{2} + e_{2}^{2} = \frac{43}{40}, e_{1} e_{2} = \frac{\sqrt{7}}{2 \sqrt{10}},$ $| e_{1}^{2} - e_{2}^{2} |$ $= \frac{27}{40}$

Q 16 :

In a triangle $A B C$ with fixed base $B C$ , the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^{2} \frac{A}{2} .$ If $a$ , $b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A$ , $B$ and $C$ , respectively, then [2009]

$b + c = 4 a$
$b + c = 2 a$
locus of point $A$ is an ellipse
locus of point $A$ is a pair of straight lines

1

2

3

4

Select one or more options

(2, 3)

$In ∆ A B C, \cos B + \cos C = 4 \sin^{2} \frac{A}{2} (Given)$

$\Rightarrow 2 \cos \frac{B + C}{2} \cos \frac{B - C}{2} - 4 \sin^{2} \frac{A}{2} = 0$

$\Rightarrow 2 \sin \frac{A}{2} [\cos \frac{B - C}{2} - 2 \sin \frac{A}{2}] = 0$

$\Rightarrow \sin \frac{A}{2} = 0 or \cos \frac{B - C}{2} - 2 \cos \frac{B + C}{2} = 0$

$Since in a triangle, \sin \frac{A}{2} \neq 0$

$∴$ $\cos \frac{B - C}{2} - 2 \cos \frac{B + C}{2} = 0$ $\Rightarrow \frac{\cos (\frac{B + C}{2})}{\cos (\frac{B - C}{2})} = \frac{1}{2}$

$By componendo and dividendo, we get$

$\frac{\cos (\frac{B + C}{2}) + \cos (\frac{B - C}{2})}{\cos (\frac{B + C}{2}) - \cos (\frac{B - C}{2})} = \frac{1 + 2}{1 - 2} = - 3$

$\Rightarrow \frac{2 \cos \frac{B}{2} \cos \frac{C}{2}}{- 2 \sin \frac{B}{2} \sin \frac{C}{2}} = - 3$ $\Rightarrow \tan \frac{B}{2} \tan \frac{C}{2} = \frac{1}{3}$

$\Rightarrow \sqrt{\frac{(s - a) (s - c)}{s (s - b)}} \sqrt{\frac{(s - a) (s - b)}{s (s - c)}} = \frac{1}{3}$

$\Rightarrow \frac{s - a}{s} = \frac{1}{3} \Rightarrow 2 s = 3 a$ $\Rightarrow a + b + c = 3 a \Rightarrow b + c = 2 a$

$i.e. A C + A B = constant$ $(∵ Base B C = a is given to be constant)$

$∴$ $Locus of A is an ellipse.$

Q 17 :

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ , where $y_{1} < 0, y_{2} < 0$ , be the end points of the latus rectum of the ellipse $x^{2} + 4 y^{2} = 4 .$ The equations of parabolas with latus rectum $P Q$ are [2008]

$x^{2} + 2 \sqrt{3} y = 3 + \sqrt{3}$
$x^{2} - 2 \sqrt{3} y = 3 + \sqrt{3}$
$x^{2} + 2 \sqrt{3} y = 3 - \sqrt{3}$
$x^{2} - 2 \sqrt{3} y = 3 - \sqrt{3}$

1

2

3

4

Select one or more options

(2, 3)

$Given ellipse is x^{2} + 4 y^{2} = 4$

$or \frac{x^{2}}{2^{2}} + \frac{y^{2}}{1} = 1 \Rightarrow a = 2, b = 1$

$∴$ $e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ $∴$ $a e = \sqrt{3}$

$As per question P \equiv (a e, - \frac{b^{2}}{a}) = (\sqrt{3}, - \frac{1}{2})$

$Q \equiv (- a e, - \frac{b^{2}}{a}) = (- \sqrt{3}, - \frac{1}{2})$ $∴$ $P Q = 2 \sqrt{3}$

$Now if P Q is the length of latus rectum of the parabola whose equation is to be found, then$

$P Q = 4 a \Rightarrow 2 \sqrt{3} = 4 a \Rightarrow a = \frac{\sqrt{3}}{2}$

Also as $P Q$ is horizontal, parabola with $P Q$ as latus rectum can be upward parabola (with vertex at A) or downward parabola (with vertex at $A'$ ) as shown in the figure.

For upward parabola,

$A R = a = \frac{\sqrt{3}}{2},$ $∴$ $Coordinates of A = (0, - (\frac{\sqrt{3} + 1}{2}))$

$∴$ $Equation of upward parabola is$

$x^{2} = 2 \sqrt{3} (y + \frac{\sqrt{3} + 1}{2})$ $\Rightarrow x^{2} - 2 \sqrt{3} y = 3 + \sqrt{3} \dots (i)$

$For downward parabola$ $A' R = a = \frac{\sqrt{3}}{2}$

$∴$ $Coordinates of A' = (0, - (\frac{1 - \sqrt{3}}{2}))$

$∴$ $Equation of downward parabola is given by$

$x^{2} = - 2 \sqrt{3} (y + \frac{1 - \sqrt{3}}{2})$ $\Rightarrow x^{2} + 2 \sqrt{3} y = 3 - \sqrt{3} \dots (i i)$

$∴$ $Equation of required parabola is given by equation (i) or (i i)$

Q 18 :

Consider the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1 .$

Let $H (α, 0)$ , $0 < α < 2$ , be a point. A straight line drawn through $H$ parallel to the $y$ -axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$ -axis at a point $G$ . Suppose the straight line joining $F$ and the origin makes an angle $ϕ$ with the positive $x$ -axis. [2022]

List-I List-II

(I) If $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is (P) $\frac{{(\sqrt{3} - 1)}^{4}}{8}$

(II) If $ϕ = \frac{π}{3}$ , then the area of the triangle $F G H$ is (Q) $1$

(III) If $ϕ = \frac{π}{6}$ , then the area of the triangle $F G H$ is (R) $\frac{3}{4}$

(IV) If $ϕ = \frac{π}{12}$ , then the area of the triangle $F G H$ is (S) $\frac{1}{2 \sqrt{3}}$

(T) $\frac{3 \sqrt{3}}{2}$

The correct option is:

(I) → (R); (II) → (S); (III) → (Q); (IV) → (P)
(I) → (R); (II) → (T); (III) → (S); (IV) → (P)
(I) → (Q); (II) → (T); (III) → (S); (IV) → (P)
(I) → (Q); (II) → (S); (III) → (Q); (IV) → (P)

1

2

3

4

(3)

$Let F (2 \cos ϕ, 2 \sin ϕ) and E (2 \cos ϕ, \sqrt{3} \sin ϕ)$

$α \equiv 2 \cos ϕ$

$Tangent at E (2 \cos ϕ, \sqrt{3} \sin ϕ) to ellipse \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$

$i.e. \frac{x \cos ϕ}{2} + \frac{y \sin ϕ}{\sqrt{3}} = 1$ $intersects x-axis at G (2 \sec ϕ, 0)$

$Area of triangle F G H = \frac{1}{2} H G \times F H$

$= \frac{1}{2} (2 \sec ϕ - 2 \cos ϕ) 2 \sin ϕ;$ $Δ = 2 \sin^{2} ϕ \cdot \tan ϕ$

$Δ = (1 - \cos 2 ϕ) \cdot \tan ϕ$

$I. If ϕ = \frac{π}{4}, Δ = 1 \to (Q)$

$II. If ϕ = \frac{π}{3}, Δ = 2 {(\frac{\sqrt{3}}{2})}^{2} \cdot \sqrt{3} = \frac{3 \sqrt{3}}{2} \to (T)$

$III. If ϕ = \frac{π}{6}, Δ = 2 {(\frac{1}{2})}^{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2 \sqrt{3}} \to (S)$

$IV. If ϕ = \frac{π}{12}, Δ = (1 - \frac{\sqrt{3}}{2}) (2 - \sqrt{3}) = \frac{{(2 - \sqrt{3})}^{2}}{2} \to (P)$

Q 19 :

Let $F_{1} (x_{1}, 0)$ and $F_{2} (x_{2}, 0)$ , for $x_{1} < 0$ and $x_{2} > 0$ , be the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1 .$ Suppose a parabola having vertex at the origin and focus at $F_{2}$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant. [2016]

Q. The orthocentre of the triangle $F_{1} M N$ is

$(- \frac{9}{10}, 0)$
$(\frac{2}{3}, 0)$
$(\frac{9}{10}, 0)$
$(\frac{2}{3}, \sqrt{6})$

1

2

3

4

(1)

$Given : Ellipse \frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$

$e = \sqrt{1 - \frac{8}{9}} = \frac{1}{3} \dots (i)$

$∴$ $F_{1} (- 1, 0) and F_{2} (1, 0)$

$Parabola with vertex at (0, 0) and focus at F_{2} (1, 0) is$

$y^{2} = 4 x \dots (i i)$

$∴$ $On solving (i) and (i i), we get the intersection points of ellipse and parabola as$

$M (\frac{3}{2}, \sqrt{6}) and N (\frac{3}{2}, - \sqrt{6})$

$One altitude of ∆ F_{1} M N is x-axis i.e. y = 0 and altitude from M to F_{1} N is$

$y - \sqrt{6} = \frac{5}{2 \sqrt{6}} (x - \frac{3}{2})$

$Putting y = 0 in above equation, we get$ $x = - \frac{9}{10}$

$∴$ $Orthocentre (- \frac{9}{10}, 0)$

Q 20 :

Let $F_{1} (x_{1}, 0)$ and $F_{2} (x_{2}, 0)$ , for $x_{1} < 0$ and $x_{2} > 0$ , be the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1 .$ Suppose a parabola having vertex at the origin and focus at $F_{2}$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant. [2016]

Q. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the $x$ -axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral $M F_{1} N F_{2}$ is

3 : 4
4 : 5
5 : 8
2 : 3

1

2

3

4

(3)

$Given : Ellipse \frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$

$e = \sqrt{1 - \frac{8}{9}} = \frac{1}{3} \dots (i)$

$∴$ $F_{1} (- 1, 0) and F_{2} (1, 0)$

$Parabola with vertex at (0, 0) and focus at F_{2} (1, 0) is$

$y^{2} = 4 x \dots (i i)$

$∴$ $On solving (i) and (i i), we get the intersection points of ellipse and parabola as$

$M (\frac{3}{2}, \sqrt{6}) and N (\frac{3}{2}, - \sqrt{6})$

$Tangents to ellipse at M and N are$

$\frac{x}{6} + \frac{y \sqrt{6}}{8} = 1 \dots (i)$

$and \frac{x}{6} - \frac{y \sqrt{6}}{8} = 1 \dots (i i)$

$On solving (i) and (i i), we get their intersection point R (6, 0) .$

$Now equation of normal to parabola at M (\frac{3}{2}, \sqrt{6}) is$

$y - \sqrt{6} = - \frac{\sqrt{6}}{2} (x - \frac{3}{2})$

$Its intersection with x-axis is Q (\frac{7}{2}, 0)$

$Now area (∆ M Q R) = \frac{1}{2} \times \frac{5}{2} \times \sqrt{6} = \frac{5 \sqrt{6}}{4}$

$Now area (M F_{1} N F_{2}) = 2 \times area (F_{1} M F_{2})$ $= 2 \times \frac{1}{2} \times 2 \times \sqrt{6} = 2 \sqrt{6}$

$∴$ $\frac{area (∆ M Q R)}{area (M F_{1} N F_{2})} = \frac{5 \sqrt{6}}{4 \times 2 \sqrt{6}} = 5 : 8$

Q 21 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The coordinates of A and B are

$(3, 0) and (0, 2)$
$(- \frac{8}{5}, \frac{2 \sqrt{161}}{15}) and (- \frac{9}{5}, \frac{8}{5})$
$(- \frac{8}{5}, \frac{2 \sqrt{161}}{15}) and (0, 2)$
$(3, 0) and (- \frac{9}{5}, \frac{8}{5})$

1

2

3

4

(4)

$Tangent to \frac{x^{2}}{3^{2}} + \frac{y^{2}}{2^{2}} = 1 at the point (3 \cos θ, 2 \sin θ) is$

$\frac{x \cos θ}{3} + \frac{y \sin θ}{2} = 1$

$Since it passes through (3, 4),$

$∴$ $\cos θ + 2 \sin θ = 1$

$\Rightarrow 4 \sin^{2} θ = 1 + \cos^{2} θ - 2 \cos θ$

$\Rightarrow 5 \cos^{2} θ - 2 \cos θ - 3 = 0$

$\Rightarrow \cos θ = 1, - \frac{3}{5}$ $\Rightarrow \sin θ = 0, \frac{4}{5}$

$∴$ $Required points are A (3, 0) and B (- \frac{9}{5}, \frac{8}{5})$

Q 22 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The orthocenter of the triangle PAB is

$(5, \frac{8}{7})$
$(\frac{7}{5}, \frac{25}{8})$
$(\frac{11}{5}, \frac{8}{5})$
$(\frac{8}{25}, \frac{7}{5})$

1

2

3

4

(3)

$Let H be the orthocentre of ∆ P A B, then as B H ⊥ A P,$ $B H is a horizontal line through B .$

$∴$ $y-coordinate of B = \frac{8}{5}$

$Let H has coordinate (α, \frac{8}{5}) .$

$∴$ $Slope of P H = \frac{\frac{8}{5} - 4}{α - 3} = \frac{- 12}{5 (α - 3)}$

$and slope of A B = \frac{\frac{8}{5} - 0}{- \frac{9}{5} - 3} = \frac{8}{- 24} = - \frac{1}{3}$

$But P H ⊥ A B,$ $∴$ $\frac{- 12}{5 (α - 3)} \times (\frac{- 1}{3}) = - 1$

$\Rightarrow 4 = - 5 α + 15 or α = \frac{11}{5}$ $∴$ $H (\frac{11}{5}, \frac{8}{5}) .$

Q 23 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The equation of the locus of the point whose distances from the point P and the line AB are equal, is

$9 x^{2} + y^{2} - 6 x y - 54 x - 62 y + 241 = 0$
$x^{2} + 9 y^{2} + 6 x y - 54 x + 62 y - 241 = 0$
$9 x^{2} + 9 y^{2} - 6 x y - 54 x - 62 y - 241 = 0$
$x^{2} + y^{2} - 2 x y + 27 x + 31 y - 120 = 0$

1

2

3

4

(1)

$Clearly the moving point traces a parabola with focus at P (3, 4) and directrix as$

$A B : \frac{y - 0}{x - 3} = - \frac{1}{3} \Rightarrow x + 3 y - 3 = 0$

$∴$ $Equation of parabola is$

${(x - 3)}^{2} + {(y - 4)}^{2} = \frac{{(x + 3 y - 3)}^{2}}{10}$

$\Rightarrow 9 x^{2} + y^{2} - 6 x y - 54 x - 62 y + 241 = 0$

Topic Question Set

Q 3 :

The ellipse $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point (0, 4) circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is [2012]

Q 4 :

The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $x$ -axis at $Q$ . If $M$ is the midpoint of the line segment $P Q$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points [2009]

Q 5 :

The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^{2} + 9 y^{2} = 9$ meets its auxiliary circle at the point $M$ . Then the area of the triangle with vertices at $A, M$ and the origin $O$ is [2009]

Q 6 :

The minimum area of triangle formed by the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and coordinate axes is [2005]

Q 7 :

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is [2004]

Q 8 :

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ , is [2003]

Q 16 :

In a triangle $A B C$ with fixed base $B C$ , the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^{2} \frac{A}{2} .$ If $a$ , $b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A$ , $B$ and $C$ , respectively, then [2009]

Q 17 :

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ , where $y_{1} < 0, y_{2} < 0$ , be the end points of the latus rectum of the ellipse $x^{2} + 4 y^{2} = 4 .$ The equations of parabolas with latus rectum $P Q$ are [2008]

Q 21 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The coordinates of A and B are

Q 22 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The orthocenter of the triangle PAB is

Q 23 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The equation of the locus of the point whose distances from the point P and the line AB are equal, is

Want Us to Email you About Special Offers & Updates?

	List-I		List-II
(I)	If $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is	(P)	$\frac{{(\sqrt{3} - 1)}^{4}}{8}$
(II)	If $ϕ = \frac{π}{3}$ , then the area of the triangle $F G H$ is	(Q)	$1$
(III)	If $ϕ = \frac{π}{6}$ , then the area of the triangle $F G H$ is	(R)	$\frac{3}{4}$
(IV)	If $ϕ = \frac{π}{12}$ , then the area of the triangle $F G H$ is	(S)	$\frac{1}{2 \sqrt{3}}$
		(T)	$\frac{3 \sqrt{3}}{2}$

Topic Question Set

Q 1 : Let S denote the locus of the point of intersection of the pair of lines 4x-3y=12α, 4αx+3αy=12, where α varies over the set of non-zero real numbers. Let T be the tangent to S passing through the points (p,0) and (0,q), q>0 and parallel to the line 4x-32y=0. Then the value of pq is [2025]

Q 3 : The ellipse E1:x29+y24=1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is [2012]

Q 4 : The normal at a point P on the ellipse x2+4y2=16 meets the x-axis at Q. If M is the midpoint of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points [2009]

Q 5 : The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2+9y2=9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A,M and the origin O is [2009]

Q 6 : The minimum area of triangle formed by the tangent to the ellipse x2a2+y2b2=1 and coordinate axes is [2005]

Q 7 : If tangents are drawn to the ellipse x2+2y2=2, then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is [2004]

Q 8 : The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse x29+y25=1, is [2003]

Q 10 : A vertical line passing through the point (h,0) intersects the ellipse x24+y23=1 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If Δ(h) is the area of the triangle PQR, Δ1=max1/2≤h≤1Δ(h) and Δ2=min1/2≤h≤1Δ(h), then 85Δ1-8Δ2= [2013]

Q 16 : In a triangle ABC with fixed base BC, the vertex A moves such that cosB+cosC=4sin2A2. If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then [2009]

Q 17 : Let P(x1,y1) and Q(x2,y2), where y1<0, y2<0, be the end points of the latus rectum of the ellipse x2+4y2=4. The equations of parabolas with latus rectum PQ are [2008]

Q 19 : Let F1(x1,0) and F2(x2,0), for x1<0 and x2>0, be the foci of the ellipse x29+y28=1. Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. [2016] Q. The orthocentre of the triangle F1MN is

Q 21 : Tangents are drawn from the point P(3, 4) to the ellipse x29+y24=1 touching the ellipse at points A and B. [2010] Q. The coordinates of A and B are

Q 22 : Tangents are drawn from the point P(3, 4) to the ellipse x29+y24=1 touching the ellipse at points A and B. [2010] Q. The orthocenter of the triangle PAB is

Q 23 : Tangents are drawn from the point P(3, 4) to the ellipse x29+y24=1 touching the ellipse at points A and B. [2010] Q. The equation of the locus of the point whose distances from the point P and the line AB are equal, is

Want Us to Email you About Special Offers & Updates?

Q 3 :

The ellipse $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point (0, 4) circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is [2012]

Q 4 :

The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $x$ -axis at $Q$ . If $M$ is the midpoint of the line segment $P Q$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points [2009]

Q 5 :

The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^{2} + 9 y^{2} = 9$ meets its auxiliary circle at the point $M$ . Then the area of the triangle with vertices at $A, M$ and the origin $O$ is [2009]

Q 6 :

The minimum area of triangle formed by the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and coordinate axes is [2005]

Q 7 :

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is [2004]

Q 8 :

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ , is [2003]

Q 16 :

In a triangle $A B C$ with fixed base $B C$ , the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^{2} \frac{A}{2} .$ If $a$ , $b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A$ , $B$ and $C$ , respectively, then [2009]

Q 17 :

Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ , where $y_{1} < 0, y_{2} < 0$ , be the end points of the latus rectum of the ellipse $x^{2} + 4 y^{2} = 4 .$ The equations of parabolas with latus rectum $P Q$ are [2008]

Q 21 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The coordinates of A and B are

Q 22 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The orthocenter of the triangle PAB is

Q 23 :

Tangents are drawn from the point P(3, 4) to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ touching the ellipse at points A and B. [2010]

Q. The equation of the locus of the point whose distances from the point P and the line AB are equal, is