For the parabola y^2 = x,let PQ be a chord su | vedclass

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(B) The equation of the parabola is $y^2 = x$.The axis of the parabola is the $x$-axis $(y = 0)$.The chord $PQ$ is perpendicular to the axis of the pa

Vedclass · Accepted Answer

$-4$

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(B) The equation of the parabola is $y^2 = x$.The axis of the parabola is the $x$-axis $(y = 0)$.The chord $PQ$ is perpendicular to the axis of the parabola,so $PQ$ is a vertical line.Since $P$ is $(4, -2)$,the equation of the line $PQ$ is $x = 4$.To find the coordinates of $Q$,we substitute $x = 4$ into the parabola equation $y^2 = x$:$y^2 = 4 \implies y = 2$ or $y = -2$.Since $P$ is $(4, -2)$,the other endpoint $Q$ must be $(4, 2)$.Now,we find the slope of the tangent at $Q(4, 2)$ by differentiating $y^2 = x$ with respect to $x$:$2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$.At $Q(4, 2)$,the slope of the tangent $m_t = \frac{1}{2(2)} = \frac{1}{4}$.The slope of the normal $m_n$ is given by $-\frac{1}{m_t}$.Therefore,$m_n = -\frac{1}{1/4} = -4$.

For the parabola $y^2 = x$,let $PQ$ be a chord such that one endpoint $P$ is $(4, -2)$ and the chord is perpendicular to the axis of the parabola. What is the slope of the normal at $Q$?

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