Find the equation of the normal to the parabo | vedclass

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(D) Step $1$: Find the upper end of the latus rectum for the given parabola.The coordinates of the ends of the latus rectum of the parabola $y^2 = -4a

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$x - y + 9 = 0$

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(D) Step $1$: Find the upper end of the latus rectum for the given parabola.The coordinates of the ends of the latus rectum of the parabola $y^2 = -4ax$ are $(-a, \pm 2a)$.Comparing with the given parabola $y^2 = -12x$,we get $4a = 12$,so $a = 3$.Hence,the upper end of its latus rectum is $(-3, 6)$.Step $2$: Find the slope of the normal to the given parabola at this point.Given equation of the parabola is $y^2 = -12x$.Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = -12$,which implies $\frac{dy}{dx} = \frac{-6}{y}$.The slope of the tangent at $(-3, 6)$ is $m_t = \frac{-6}{6} = -1$.The slope of the normal at $(-3, 6)$ is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.Step $3$: Find the equation of the normal.The equation of the normal is a line passing through $(-3, 6)$ with slope $m = 1$.Using the point-slope form $y - y_1 = m(x - x_1)$:$y - 6 = 1(x - (-3))$$y - 6 = x + 3$$x - y + 9 = 0$.

Find the equation of the normal to the parabola $y^2 + 12x = 0$ at the upper end of its latus rectum.

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