Chord PQ of parabola y^2 = x, where one end P of chord is at point (4,-2) is perpendicular to axis of parabola. Then slope of normal at Q is?

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$$\begin{gathered} \mathrm{Chord} \mathrm{PQ} \mathrm{of} \mathrm{parabola} \mathrm{y}^2 = \mathrm{x}, \mathrm{where} \mathrm{one} \mathrm{end} \mathrm{P} \mathrm{of} \mathrm{chord} \mathrm{is} \mathrm{at} \mathrm{point} \\ (4,-2) \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{axis} \mathrm{of} \mathrm{parabola}. \mathrm{Then} \mathrm{slope} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{Q} \mathrm{is} \\ \mathrm{We} \mathrm{have} \\ {\mathrm{y}}^2=\mathrm{x} \\ \mathrm{As} \mathrm{PQ} \mathrm{is} \mathrm{perpedicular} \mathrm{to} \mathrm{axis} \mathrm{of} \mathrm{parabola}, \mathrm{hence} \mathrm{PQ}\hspace{0.17em}\mathrm{is} \mathrm{double} \\ \mathrm{ordinate} \\ \mathrm{Hence} \mathrm{coordinate} \mathrm{of} \mathrm{P}\hspace{0.17em}\mathrm{is} \left(4,-2\right) \mathrm{means} \mathrm{coordinate} \mathrm{of} \mathrm{Q} \mathrm{is} \left(4,2\right) \\ {\mathrm{y}}^2=\mathrm{x} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ 2\mathrm{y}.\frac{\mathrm{dy}}{\mathrm{dx}}=1 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\mathrm{y}} \\ \mathrm{Slope} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{Q}=-{\frac{\mathrm{dx}}{\mathrm{dy}}}_{\left(4,2\right)} \\ ={\left(-2\mathrm{y}\right)}_{\left(4,2\right)} \\ =-2\times 2 \\ =-4 \left(\mathrm{Answer}\right) \end{gathered}$$

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