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 $(a, -2a)$.For the parabola $y^{2}=4ax$,the slope of

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

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(A) The coordinates of the ends of the latus rectum of the parabola $y^{2}=4ax$ are $(a, 2a)$ and $(a, -2a)$.For the parabola $y^{2}=4ax$,the slope of the tangent at $(x_{1}, y_{1})$ is $m = \frac{2a}{y_{1}}$.At $(a, 2a)$,the slope of the tangent is $m = \frac{2a}{2a} = 1$,so the slope of the normal is $-1$.The equation of the normal at $(a, 2a)$ is $y - 2a = -1(x - a)$,which simplifies to $x + y - 3a = 0$.At $(a, -2a)$,the slope of the tangent is $m = \frac{2a}{-2a} = -1$,so the slope of the normal is $1$.The equation of the normal at $(a, -2a)$ is $y - (-2a) = 1(x - a)$,which simplifies to $x - y - 3a = 0$.The combined equation is $(x + y - 3a)(x - y - 3a) = 0$.Expanding this,we get $((x - 3a) + y)((x - 3a) - y) = 0$,which is $(x - 3a)^{2} - y^{2} = 0$.Thus,$x^{2} - 6ax + 9a^{2} - y^{2} = 0$,or $x^{2} - y^{2} - 6ax + 9a^{2} = 0$.

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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