JEE Main 2013 (Online) 25th AprilQuestion 1
A common tangent to the conics $x^{2}=6y$ and $2x^{2}-4y^{2}=9$ is :
Hyperbola
Solution
Tangent to $x^2=6y$: $y=mx-\frac32m^2$. For $\frac{x^2}{9/2}-\frac{y^2}{9/4}=1$ need $c^2=\frac92m^2-\frac94$: $\frac94m^4=\frac92m^2-\frac94$, so $m^2=1$, $c=-\frac32$. Tangent $x-y=\frac32$.