KCET · Maths · Vector Algebra
Slope of Normal to the curve \( y=x^{2}-\frac{1}{x^{2}} \) at \( (-1,0) \) is
- A \( \frac{1}{4} \)
- B \( 0-\frac{1}{4} \)
- C 4
- D \(-4 \)
Answer & Solution
Correct Answer
(A) \( \frac{1}{4} \)
Step-by-step Solution
Detailed explanation
Given equation of curve,
\[
\begin{array}{l}
y=x^{2}-\frac{1}{x^{2}} \\
\Rightarrow y=x^{2}-x^{-2}
\end{array}
\]
Differentiating both the sides, we get
\[
y^{\prime}=2 x-(-2) x^{-3}=2 x+\frac{2}{x^{3}}
\]
\[
\text { At }(-1,0), \text { we get } m_{1}=-2-2=-4
\]
So, slope of the curve is \( -4 \)
We know that, slopes of two perpendicular lines is given by
\[
\begin{array}{l}
m_{1} m_{2}=-1 \\
\Rightarrow m_{2}=\frac{1}{4}
\end{array}
\]
Hence, the slope of the normal to the curve is \( \frac{1}{4} \)
\[
\begin{array}{l}
y=x^{2}-\frac{1}{x^{2}} \\
\Rightarrow y=x^{2}-x^{-2}
\end{array}
\]
Differentiating both the sides, we get
\[
y^{\prime}=2 x-(-2) x^{-3}=2 x+\frac{2}{x^{3}}
\]
\[
\text { At }(-1,0), \text { we get } m_{1}=-2-2=-4
\]
So, slope of the curve is \( -4 \)
We know that, slopes of two perpendicular lines is given by
\[
\begin{array}{l}
m_{1} m_{2}=-1 \\
\Rightarrow m_{2}=\frac{1}{4}
\end{array}
\]
Hence, the slope of the normal to the curve is \( \frac{1}{4} \)
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