The point of intersection of tangents at the | vedclass

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(B) The parabola is given by $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.The ends of the latus rectum are $L(a, 2a)$ and $L_1(a, -2a)$,

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$(-1,0)$

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(B) The parabola is given by $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.The ends of the latus rectum are $L(a, 2a)$ and $L_1(a, -2a)$,which are $L(1, 2)$ and $L_1(1, -2)$.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 point $L(1, 2)$,the tangent equation is $y(2) = 2(1)(x + 1) \implies y = x + 1$.For point $L_1(1, -2)$,the tangent equation is $y(-2) = 2(1)(x + 1) \implies -y = x + 1$.Adding the two equations: $(x + 1) + (x + 1) = 0 \implies 2x + 2 = 0 \implies x = -1$.Substituting $x = -1$ into $y = x + 1$,we get $y = -1 + 1 = 0$.Thus,the point of intersection is $(-1, 0)$.

The point of intersection of tangents at the ends of the latus rectum of the parabola $y^2 = 4x$ is equal to

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