The locus of the midpoint of the chord of the | vedclass

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(A) Let the midpoint of the chord be $(x_1, y_1)$.The equation of the chord of the parabola $y^2 = 4ax$ with midpoint $(x_1, y_1)$ is given by $T = S_

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$y^2 - 2ax + 8a^2 = 0$

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(A) Let the midpoint of the chord be $(x_1, y_1)$.The equation of the chord of the parabola $y^2 = 4ax$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,which is $yy_1 - 2a(x + x_1) = y_1^2 - 4ax_1$.This simplifies to $yy_1 - 2ax = y_1^2 - 2ax_1$,or $\frac{yy_1 - 2ax}{y_1^2 - 2ax_1} = 1$.To find the lines joining the vertex $(0, 0)$ to the intersection points of the chord and the parabola,we homogenize the equation of the parabola $y^2 - 4ax = 0$ using the chord equation:$y^2 - 4ax(1) = 0 \implies y^2 - 4ax \left( \frac{yy_1 - 2ax}{y_1^2 - 2ax_1} \right) = 0$.Expanding this,we get $y^2(y_1^2 - 2ax_1) - 4axyy_1 + 8a^2x^2 = 0$.Since the chord subtends a right angle at the vertex,the lines are perpendicular,meaning the sum of the coefficients of $x^2$ and $y^2$ must be zero:$8a^2 + (y_1^2 - 2ax_1) = 0$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $y^2 - 2ax + 8a^2 = 0$.

The locus of the midpoint of the chord of the parabola $y^2 = 4ax$ which subtends a right angle at the vertex is

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