The equations of the normals at the ends of t | vedclass

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(A) The coordinates of the ends of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.The equation of the normal at $(x_1, y_1)

Vedclass · Accepted Answer

$x^2 - y^2 - 6ax + 9a^2 = 0$

Vedclass · Answer

(A) The coordinates of the ends of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.The equation of the normal at $(x_1, y_1)$ to $y^2 = 4ax$ is given by $y - y_1 = -\frac{y_1}{2a}(x - x_1)$.For the point $(a, 2a)$,the normal is $y - 2a = -\frac{2a}{2a}(x - a) \implies y - 2a = -(x - a) \implies x + y - 3a = 0$ $(i)$.For the point $(a, -2a)$,the normal is $y - (-2a) = -\frac{-2a}{2a}(x - a) \implies y + 2a = 1(x - a) \implies x - y - 3a = 0$ $(ii)$.The combined equation is $(x + y - 3a)(x - y - 3a) = 0$.This simplifies to $((x - 3a) + y)((x - 3a) - y) = 0 \implies (x - 3a)^2 - y^2 = 0$.Expanding this,we get $x^2 - 6ax + 9a^2 - y^2 = 0$,which is $x^2 - y^2 - 6ax + 9a^2 = 0$.

The equations of the normals at the ends of the latus rectum of the parabola $y^2 = 4ax$ are given by

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