Question

Consider the parabola $y^2=20x$, ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and hyperbola $\frac{x^2}{29}-\frac{y^2}{4}=1$
Area of the quadrilateral formed by common tangents and chord of contacts of ellipse and parabola respectively, is

• A. $\frac{2149}{25}$
• B. $\frac{2191}{25}$
• C. $\frac{2194}{25}$
• D. $\frac{2419}{25}$

Answer 1

Answer: (D)

Point of intersection of common tangents is $P(-5,0)$.
Chord of contact from $P$ to the parabola is
$0\cdot y=10(x-5)\Rightarrow x=5$
to the ellipse is $\frac{x(-5)}{16}+0=1\Rightarrow x=\frac{-16}{5}$
So, the quadrilateral formed is trapezium.
$\text{ Required area }=\frac{1}{2}\left(\frac{41}{5}\right)\left(20+\frac{18}{5}\right)$
$=\frac{1}{2}\left(\frac{41}{5}\right)\left(\frac{118}{5}\right)=\frac{(41)(59)}{25}$