Let A ( x 1, 0) and B ( x 2, 0), for x 10 be the foci of the hyperbola ( x 2/9)-( y 2/16)=1. Suppose a parabola having vertex at origin and focus at B intersect the hyperbola at P in first quadrant and at point Q in fourth quadrant. If an ellipse passes through foci of hyperbola with foci of ellipse at vertices of hyperbola, then area of the quadrilateral formed by tangents at the ends of latera recta of ellipse is equal to

Question

Let A ( x 1, 0) and B ( x 2, 0), for x 1<0 x 2>0 be the foci of the hyperbola ( x 2/9)-( y 2/16)=1. Suppose a parabola having vertex at origin and focus at B intersect the hyperbola at P in first quadrant and at point Q in fourth quadrant. If an ellipse passes through foci of hyperbola with foci of ellipse at vertices of hyperbola, then area of the quadrilateral formed by tangents at the ends of latera recta of ellipse is equal to (A) (250/3) (B) (125/3) (C) (50/3) (D) (25/3)

Tardigrade Admin · Accepted Answer

Foci of ellipse =( ± 3,0) and vertices =( ± 5,0) ∴ Ellipse is ( x 2/25)+( y 2/16)=1 ∴ Area of quadrilateral =(2 a 2/ e )=2 × (25/(3)5)=(250/3).

$\frac{250}{3}$

$\frac{125}{3}$

$\frac{50}{3}$

$\frac{25}{3}$

Foci of ellipse $= (\pm 3, 0)$ and vertices $= (\pm 5, 0)$
$∴$ Ellipse is $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$
$∴$ Area of quadrilateral $= \frac{2 a ^{2}}{e} = 2 \times \frac{25}{\frac{3}{5}} = \frac{250}{3}$ .

3250​

3125​

350​

325​

Foci of ellipse =(±3,0) and vertices =(±5,0) ∴ Ellipse is 25x2​+16y2​=1 ∴ Area of quadrilateral =e2a2​=2×53​25​=3250​.

$\frac{250}{3}$

$\frac{125}{3}$

$\frac{50}{3}$

$\frac{25}{3}$

Foci of ellipse $= (\pm 3, 0)$ and vertices $= (\pm 5, 0)$
$∴$ Ellipse is $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$
$∴$ Area of quadrilateral $= \frac{2 a ^{2}}{e} = 2 \times \frac{25}{\frac{3}{5}} = \frac{250}{3}$ .