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(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $t$ is $ty = x + at^2$.Let the tangents at $t_1$ and $t_2$ be $t_1y = x + at_1^2$

Vedclass · Accepted Answer

$(at_1t_2, a(t_1 + t_2))$

Vedclass · Answer

(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $t$ is $ty = x + at^2$.Let the tangents at $t_1$ and $t_2$ be $t_1y = x + at_1^2$ and $t_2y = x + at_2^2$.Subtracting the two equations: $(t_1 - t_2)y = a(t_1^2 - t_2^2) = a(t_1 - t_2)(t_1 + t_2)$.Thus,$y = a(t_1 + t_2)$.Substituting $y$ into the first equation: $t_1(a(t_1 + t_2)) = x + at_1^2$.$at_1^2 + at_1t_2 = x + at_1^2$,which gives $x = at_1t_2$.Therefore,the point of intersection is $(at_1t_2, a(t_1 + t_2))$.

The point of intersection of the tangents to the parabola $y^2 = 4ax$ at the points $t_1$ and $t_2$ is

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