If two normals to a parabola y^2 = 4ax inters | vedclass

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(B) Let the two normals to the parabola $y^2 = 4ax$ be at points $t_1$ and $t_2$. The equation of a normal at point $t$ is $y + tx = 2at + at^3$.If th

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$(a, 0)$

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(B) Let the two normals to the parabola $y^2 = 4ax$ be at points $t_1$ and $t_2$. The equation of a normal at point $t$ is $y + tx = 2at + at^3$.If the normals intersect at $(h, k)$,then $at^3 + (2a - h)t + k = 0$. Let the roots be $t_1, t_2, t_3$.Given that the normals are at right angles,the product of their slopes $m_1 m_2 = -1$. Since $m = -t$,we have $(-t_1)(-t_2) = -1$,so $t_1 t_2 = -1$.The equation of the chord joining $t_1$ and $t_2$ is $y(t_1 + t_2) = 2x + 2a t_1 t_2$.Substituting $t_1 t_2 = -1$,we get $y(t_1 + t_2) = 2x - 2a$,or $2x - 2a - y(t_1 + t_2) = 0$.For this line to pass through a fixed point independent of $t_1$ and $t_2$,the coefficient of $(t_1 + t_2)$ must be zero,so $y = 0$.Substituting $y = 0$ into the equation,we get $2x - 2a = 0$,which gives $x = a$.Thus,the fixed point is $(a, 0)$.

If two normals to a parabola $y^2 = 4ax$ intersect at right angles,then the chord joining their feet passes through a fixed point whose coordinates are:

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