The chord PQ of the parabola y^2 = x,where on | vedclass

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(A) The equation of the parabola is $y^2 = x$.Given that the chord $PQ$ is perpendicular to the axis of the parabola (the $x$-axis) and one end $P$ is

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$-4$

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(A) The equation of the parabola is $y^2 = x$.Given that the chord $PQ$ is perpendicular to the axis of the parabola (the $x$-axis) and one end $P$ is $(4, -2)$,the $x$-coordinate of $Q$ must also be $4$.Since $Q$ lies on the parabola $y^2 = x$,substituting $x = 4$ gives $y^2 = 4$,so $y = \pm 2$. Since $P$ is $(4, -2)$,$Q$ must be $(4, 2)$.The derivative of the parabola $y^2 = x$ with respect to $x$ is $2y \frac{dy}{dx} = 1$,which gives $\frac{dy}{dx} = \frac{1}{2y}$.The slope of the tangent at $Q(4, 2)$ is $\frac{dy}{dx} \Big|_{(4, 2)} = \frac{1}{2(2)} = \frac{1}{4}$.The slope of the normal at $Q$ is the negative reciprocal of the slope of the tangent.Therefore,the slope of the normal is $-\frac{1}{1/4} = -4$.

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

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