Variable chords of the parabola y^2 = 4ax sub | vedclass

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(D) Let the chord be $y = mx + c$. Since it passes through the parabola $y^2 = 4ax$,the combined equation of the lines joining the vertex $(0, 0)$ to

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all of the above

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(D) Let the chord be $y = mx + c$. Since it passes through the parabola $y^2 = 4ax$,the combined equation of the lines joining the vertex $(0, 0)$ to the points of intersection is $y^2 = 4ax \left(\frac{y-mx}{c}\right)$.Since the chord subtends a right angle at the vertex,the sum of the coefficients of $x^2$ and $y^2$ is $0$.$1 + \frac{4am}{c} = 0 \implies c = -4am$.Thus,the chord is $y = mx - 4am$,which always passes through the fixed point $(4a, 0)$.The locus of the feet of the perpendiculars from the vertex to the chord $y = mx - 4am$ is $x^2 + y^2 - 4ax = 0$,which is a circle.The locus of the midpoints of the chords is $y^2 = 2a(x - 4a)$,which is a parabola.Therefore,all the given statements are correct.

Variable chords of the parabola $y^2 = 4ax$ subtend a right angle at the vertex. Then:

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