The point of contact of the tangent to the pa | vedclass

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Question

(A) The slope of the tangent $m = 	an(60^\circ) = \sqrt{3}$.The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$ is $yy_

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$\left( \frac{a}{3}, \frac{2a}{\sqrt{3}} \right)$

Vedclass · Answer

(A) The slope of the tangent $m = 	an(60^\circ) = \sqrt{3}$.The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.The slope of this tangent is $m = \frac{2a}{y_1}$.Equating the slopes: $\sqrt{3} = \frac{2a}{y_1} \implies y_1 = \frac{2a}{\sqrt{3}}$.Since the point $(x_1, y_1)$ lies on the parabola $y^2 = 4ax$,we have $y_1^2 = 4ax_1$.Substituting $y_1$: $\left( \frac{2a}{\sqrt{3}} ight)^2 = 4ax_1 \implies \frac{4a^2}{3} = 4ax_1 \implies x_1 = \frac{a}{3}$.Thus,the point of contact is $\left( \frac{a}{3}, \frac{2a}{\sqrt{3}} ight)$.

The point of contact of the tangent to the parabola $y^2 = 4ax$ which makes an angle of $60^\circ$ with the $x$-axis is:

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