Match the points on the curve $2 y^2=x+1$ with the slopes of normals at those points and choose the correct…
Match the points on the curve $2 y^2=x+1$ with the slopes of normals at those points and choose the correct answer.
$\begin{array}{llll}\text { A } & \text { B } & \text { C } & \text { D } \\ 2 & 4 & 3 & 1\end{array}$
$\begin{array}{llll}\text { A } & \text { B } & \text { C } & \text { D } \\ 2 & 5 & 3 & 1\end{array}$
$\begin{array}{llll}\text { A } & \text { B } & \text { C } & \text { D } \\ 2 & 3 & 5 & 1\end{array}$
$\begin{array}{llll}\text { A } & \text { B } & \text { C } & \text { D } \\ 2 & 5 & 1 & 3\end{array}$
Solution
Given equation of curve is
$
2 y^2=x+1
$
On differentiating both sides w.r.t. $x$, we get
$
\text { 4y } \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=\frac{1}{4 y}
$
$\therefore$ The slope of the normal is
$
\frac{d x}{d y}=-4 y
$
I. $\frac{d x}{d y_{(7,2)}}=-4(2)=-8$
II. $\frac{d x}{d y_{\left(0, \frac{1}{\sqrt{2}}\right)}}=\frac{-4}{\sqrt{2}}=-2 \sqrt{2}$
III. $\frac{d x}{d y_{(1,-1)}}=-4(-1)=4$
IV. $\frac{d x}{d y_{(3, \sqrt{2})}}=-4(\sqrt{2})=-4 \sqrt{2}$
$\therefore$ Option (2) satisfy the all four statements
