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 [2016]

Question

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

Smart Achievers · Accepted Answer

(3)

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

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

$∴$ $F_{1} (- 1, 0) and F_{2} (1, 0)$

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

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

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

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

$Tangents to ellipse at M and N are$

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

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

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

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

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

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

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

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

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

Solution

1 3 : 4

2 4 : 5

3 5 : 8

4 2 : 3

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