Q.
Consider the parabola $y^2=20 x$, 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
Conic Sections
Solution:
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) $
$=\frac{1}{2}\left(\frac{41}{5}\right)\left(\frac{118}{5}\right)=\frac{(41)(59)}{25}$
