The tangent to the parabola y^2 = 4ax at the | vedclass

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(B) The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.Substituting the point $(a, 2a)

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$\frac{\pi}{4}$

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(B) The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.Substituting the point $(a, 2a)$,we get $y(2a) = 2a(x + a)$.Dividing both sides by $2a$,we obtain $y = x + a$.Comparing this with the slope-intercept form $y = mx + c$,the slope $m = 1$.Since $m = 	an 	heta$,we have $	an 	heta = 1$,which implies $	heta = \frac{\pi}{4}$.

The tangent to the parabola $y^2 = 4ax$ at the point $(a, 2a)$ makes an angle with the $x$-axis equal to

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