If line $x+y=0$ touches the curve $a x^{2}=2 y^{2}-b$ at $(1,-1)$, then the values of $a$ and $b$ are…
If line $x+y=0$ touches the curve $a x^{2}=2 y^{2}-b$ at $(1,-1)$,
then the values of $a$ and $b$ are respectively
$0,2$
$-2,0$
$0,-2$
$2,0$
Solution
$\begin{array}{l}
a x^{2}=2 y^{2}-b \\
a \times 2 x=4 y \frac{d y}{d x}-0 \Rightarrow a x=2 y \frac{d y}{d x} \\
\therefore \frac{d y}{d x}=\frac{a x}{2 y}
\end{array}$
At $(1,-1)$ slope of tangent $=\frac{d y}{d x}=\frac{-a}{2}$ and slope of $x+y=0$ is $-1$
As per condition given, $\frac{-a}{2}=-1 \Rightarrow a=2$
Substituting $x=1, y=-1, a=2$ in the given equation of curve, we get
$2=2-b \Rightarrow b=0$