If the ellipse $4 x^2+9 y^2=36$ is confocal with a hyperbola whose length of the transverse axis is 2 , then…
If the ellipse $4 x^2+9 y^2=36$ is confocal with a hyperbola whose length of the transverse axis is 2 , then the points of intersection of the ellipse and hyperbola lie on the circle
$x^2+y^2=81$
$x^2+y^2=16$
$x^2+y^2=25$
$x^2+y^2=5$
Solution
Ellipse : $4 x^2+9 y^2=36$ ....(i)
and Hyperbola $=\frac{x^2}{1}-\frac{y^2}{b^2}=1$
$\Rightarrow b^2 x^2-y^2=b^2$
As, they are confocal
$\therefore(a e)^2=a^2+b^2 \Rightarrow 5=1+b^2 \Rightarrow b^2=4$
So, hyperbola $4 x^2-y^2=4$ .....(ii)
Adding (i) and (ii), $8 x^2+8 y^2=40$
$\Rightarrow x^2+y^2=5$ intersection lies on this circle.