The equation of the normals at the ends of th | 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 slope of the tangent at $(a, 2a)$ i

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 slope of the tangent at $(a, 2a)$ is given by $2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}$. At $(a, 2a)$,the slope of the tangent is $\frac{2a}{2a} = 1$.Therefore,the slope of the normal at $(a, 2a)$ is $-1$.The equation of the normal at $(a, 2a)$ is $y - 2a = -1(x - a)$,which simplifies to $x + y - 3a = 0$ $(i)$.Similarly,the slope of the normal at $(a, -2a)$ is $1$,and the equation of the normal is $y - (-2a) = 1(x - a)$,which simplifies to $x - y - 3a = 0$ $(ii)$.The combined equation of the normals is $(x + y - 3a)(x - y - 3a) = 0$.This expands to $(x - 3a)^{2} - y^{2} = 0$,which is $x^{2} - 6ax + 9a^{2} - y^{2} = 0$ or $x^{2} - y^{2} - 6ax + 9a^{2} = 0$.

The equation of the normals at the ends of the latus rectum of the parabola $y^{2} = 4ax$ is:

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