Question

The chord PQ 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

• A. $-4$
• B. $-\frac{1}{4}$
• C. $4$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

Point P is $(4,-2)$ and PQ $\perp$ x-axis
So, $Q=(4,2)$
Equation of tangent at $(4,2)$ is
$y{y}_1=\frac{1}{2}\left(x+x_1\right)$
$\Rightarrow 2y=\frac{1}{2}\left(x+2\right)\Rightarrow 4y=x+2$
$\Rightarrow y=\frac{x}{4}+\frac{1}{2}$
So, slope of tangent $=\frac{1}{4}$
$\therefore$ Slope of normal $=-4$