Find the point where the normal drawn from th | vedclass

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(B) The given equation of the parabola is $y^2 = -12x$. Comparing this with $y^2 = -4ax$,we get $4a = 12$,so $a = 3$.The coordinates of the focus are

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$(-9, 0)$

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(B) The given equation of the parabola is $y^2 = -12x$. Comparing this with $y^2 = -4ax$,we get $4a = 12$,so $a = 3$.The coordinates of the focus are $(-a, 0) = (-3, 0)$.The latus rectum is the line $x = -3$. The upper end of the latus rectum is $(-3, 6)$.The slope of the tangent at any point $(x_1, y_1)$ on the parabola $y^2 = -4ax$ is given by $yy_1 = -2a(x + x_1)$.For the point $(-3, 6)$,the slope of the tangent $m_t$ is found by differentiating $y^2 = -12x$ with respect to $x$: $2y \frac{dy}{dx} = -12$,so $\frac{dy}{dx} = \frac{-6}{y}$.At $(-3, 6)$,the slope of the tangent $m_t = \frac{-6}{6} = -1$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.The equation of the normal at $(-3, 6)$ is $y - 6 = 1(x - (-3))$,which simplifies to $y - 6 = x + 3$,or $y = x + 9$.To find where it intersects the axis (the $x$-axis),set $y = 0$: $0 = x + 9$,which gives $x = -9$.Thus,the point of intersection is $(-9, 0)$.

Find the point where the normal drawn from the upper end of the latus rectum of the parabola $y^2 = -12x$ intersects the axis.

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