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(D) The end points of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.The equation of the tangent to the parabola $y^2 = 4ax

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The end points of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.For the point $(a, 2a)$,the tangent is $y(2a) = 2a(x + a)$,which simplifies to $y = x + a$. The slope $m_1 = 1$.For the point $(a, -2a)$,the tangent is $y(-2a) = 2a(x + a)$,which simplifies to $y = -(x + a)$. The slope $m_2 = -1$.Since $m_1 	imes m_2 = 1 	imes (-1) = -1$,the tangents are perpendicular to each other.Therefore,the angle between them is $\frac{\pi}{2}$.

The angle between the tangents drawn at the end points of the latus rectum of the parabola $y^2 = 4ax$ is

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