Explanation:
C Let the point be \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\)
\(\therefore \quad \mathrm{y}_1^2=\mathrm{x}_1\)
Now, we have
\(\mathrm{y}^2=\mathrm{x}\)
\(\therefore \quad 2 \mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=1\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}\)
\(\therefore\) Slope of normal at \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=\frac{-1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{x}_1, \mathrm{y}_1\right)}}=-\frac{1}{\left(\frac{1}{2 \mathrm{y}_1}\right)}\)
\(=-2 \mathrm{y}_1\)
It is given that,
\(-2 \mathrm{y}_1=\mathrm{x}_1\)
From equation (i) and (ii), we have-
\(\mathrm{y}_1^2=-2 \mathrm{y}_1\)
\(\mathrm{y}_1^2+2 \mathrm{y}_1=0\)
\(\mathrm{y}_1\left(\mathrm{y}_1+2\right)=0\)
\(\mathrm{y}_1=0,-2\)
\(\mathrm{x}_1=0,4\)\(\therefore \quad \mathrm{x}_1=0,4\) Required points are \((0,0),(4,-2)\).