Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 41 :

Let each of the two ellipses $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ and $E_{2} : \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1, (A < B)$ have eccentricity $\frac{4}{5}$ . Let the lengths of the latus recta of $E_{1}$ and $E_{2}$ be $l_{1}$ and $l_{2}$ , respectively, such that $2 l_{1}^{2} = 9 l_{2} .$ If the distance between the foci of $E_{1}$ is 8, then the distance between the foci of $E_{2}$ is [2026]

$\frac{8}{5}$
$\frac{96}{5}$
$\frac{16}{5}$
$\frac{32}{5}$

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Q 42 :

An ellipse has its center at (1, −2), one focus at (3, −2) and one vertex at (5, −2). Then the length of its latus rectum is : [2026]

$\frac{16}{\sqrt{3}}$
$6 \sqrt{3}$
$4 \sqrt{3}$
6

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Q 43 :

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

256
516
276
496

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Q 44 :

If the points of intersection of the ellipses $x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0$ and $4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$ , then the value of $a b + 18 r^{2}$ is. [2026]

53
51
52
55

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$By family of curve equation of circle will be$

$\Rightarrow S_{1} + λ S_{2} = 0$

$\Rightarrow (x^{2} + 2 y^{2} - 6 x - 12 y + 23) + λ (4 x^{2} + 2 y^{2} - 20 x - 12 y + 35) = 0$

$\Rightarrow for circle coeff of x^{2} = coeff of y^{2}$

$\Rightarrow λ = \frac{1}{2}$

$So equation of circle is$

$\Rightarrow x^{2} + y^{2} - \frac{16}{3} x - 6 y + \frac{27}{2} = 0$

$Centre (\frac{8}{3}, 3), Radius r = \sqrt{\frac{47}{18}}$

$∴$ $a b + 18 r^{2} = 8 + 47 = 55$

Q 45 :

If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

$\frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{7}}$
$\frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{5}}$
$\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{11}}$
$\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{5}}$

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$α x + 4 y - \sqrt{7} = 0 touches 3 x^{2} + 4 y^{2} = 1$

$∴$ $c^{2} = a^{2} m^{2} + b^{2}$

$\frac{7}{16} = \frac{1}{3} \times \frac{α^{2}}{16} + \frac{1}{4} \Rightarrow α = 3, - 3$

$Tangent is 3 x + 4 y - \sqrt{7} = 0$

Let the point of contact be $P (x_{1}, y_{1})$

$∴$ $Tangent is 3 x x_{1} + 4 y y_{1} = 1$

$∴$ $\frac{3 x_{1}}{3} = \frac{4 y_{1}}{4} = \frac{1}{\sqrt{7}}$ $∴$ $P (\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}})$

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

$P S = e (P M)$

$= e (\frac{a}{e} - \frac{1}{\sqrt{7}})$

$= \frac{1}{2} (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{7}}) = \frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{7}}$

$P S' = e (P M')$

$= \frac{1}{2} (\frac{a}{e} + \frac{1}{\sqrt{7}}) = \frac{1}{2} (\frac{1}{\sqrt{7}} + \frac{2}{\sqrt{3}})$

$= \frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{7}}$

Q 46 :

Let S and S′ be the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If ${(S P)}^{2} + {(S' P)}^{2} - S P \cdot S' P = 37,$ then $α^{2} + β^{2}$ is equal to: [2026]

13
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11

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$∵$ $P lies on ellipse \Rightarrow \frac{α^{2}}{25} + \frac{β^{2}}{9} = 1$

$∵$ $P S + P S' = 2 a \Rightarrow P S + P S' = 10$

$∴$ ${(P S)}^{2} + {(P S')}^{2} - P S \cdot P S' = 37$

${(P S + P S')}^{2} - 3 P S \cdot P S' = 37$

$100 - 3 P S \cdot P S' = 37$

$3 P S \cdot P S' = 63 \Rightarrow P S \cdot P S' = 21$

$∵$ $P S & P S' are (5 \pm \frac{4}{5} α)$

$∴$ $P S \cdot P S' = 25 - \frac{16}{25} α^{2} = 21$

$\frac{16}{25} α^{2} = 4$

$α = \frac{5}{2} \Rightarrow α^{2} = \frac{25}{4}$

$∴$ $β^{2} = \frac{27}{4}$

$∴$ $α^{2} + β^{2} = \frac{52}{4} = 13$

Topic Question Set

Q 42 :

An ellipse has its center at (1, −2), one focus at (3, −2) and one vertex at (5, −2). Then the length of its latus rectum is : [2026]

Q 43 :

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

Q 44 :

If the points of intersection of the ellipses $x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0$ and $4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$ , then the value of $a b + 18 r^{2}$ is. [2026]

Q 45 :

If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

Q 46 :

Let S and S′ be the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If ${(S P)}^{2} + {(S' P)}^{2} - S P \cdot S' P = 37,$ then $α^{2} + β^{2}$ is equal to: [2026]

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