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MHT CET · Maths · Application of Derivatives

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

  1. A \(0,2\)
  2. B \(-2,0\)
  3. C \(0,-2\)
  4. D \(2,0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2,0\)

Step-by-step Solution

Detailed explanation

\(
\begin{array}{l}
a x^{2}=2 y^{2}-b \\
a \times 2 x=4 y \frac{d y}{d x}-0 \Rightarrow a x=2 y \frac{d y}{d x} \\
\therefore \frac{d y}{d x}=\frac{a x}{2 y}
\end{array}
\)
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
\(
2=2-b \Rightarrow b=0
\)