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
- A \(0,2\)
- B \(-2,0\)
- C \(0,-2\)
- D \(2,0\)
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
\(
2=2-b \Rightarrow b=0
\)
\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
\(
2=2-b \Rightarrow b=0
\)
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