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

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Aicharya · Accepted Answer

Point $P$ is $(4,-2)$ and $P Q \perp x$-axis So, $Q=(4,2)$

Equation of tangent at $(4,2)$ is $ \begin{aligned} & y y_1=\frac{1}{2}\left(x+x_1\right) \\ & \Rightarrow 2 y=\frac{1}{2}(x+2) \Rightarrow 4 y=x+2 \\ & \Rightarrow y=\frac{x}{4}+\frac{1}{2} \end{aligned} $ So, slope of tangent $=\frac{1}{4}$ $\therefore$ Slope of normal $=-4$

The chord $P Q$ of the parabola $y^2=x$, where one end $P$ of the chord is at point $(4,-2)$, is…

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