The locus of the feet of the perpendiculars d | vedclass

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(A) Let the chord be $y = mx + c$. The intersection points of the line and the parabola $y^2 = 4ax$ are given by $(mx + c)^2 = 4ax$,which simplifies t

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$x^2 + y^2 - 4ax = 0$

Vedclass · Answer

(A) Let the chord be $y = mx + c$. The intersection points of the line and the parabola $y^2 = 4ax$ are given by $(mx + c)^2 = 4ax$,which simplifies to $m^2x^2 + (2mc - 4a)x + c^2 = 0$.Let the points be $A(x_1, y_1)$ and $B(x_2, y_2)$. Then $x_1x_2 = \frac{c^2}{m^2}$ and $y_1y_2 = \frac{4ac}{m}$.Since the chord subtends a right angle at the vertex $(0,0)$,the product of the slopes of $OA$ and $OB$ is $-1$,so $\frac{y_1}{x_1} 	imes \frac{y_2}{x_2} = -1$,which implies $y_1y_2 = -x_1x_2$.Substituting the values,$\frac{4ac}{m} = -\frac{c^2}{m^2}$,which gives $c = -4am$.The equation of the chord is $y = mx - 4am$.Let $(h, k)$ be the foot of the perpendicular from the origin to this line. The slope of the perpendicular line is $-\frac{1}{m}$,so $\frac{k}{h} = -\frac{1}{m}$,which gives $m = -\frac{h}{k}$.Since $(h, k)$ lies on the line $y = mx - 4am$,we have $k = mh - 4am = m(h - 4a)$.Substituting $m = -\frac{h}{k}$,we get $k = -\frac{h}{k}(h - 4a)$,which simplifies to $k^2 = -h^2 + 4ah$.Thus,$h^2 + k^2 - 4ah = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 4ax = 0$.

The locus of the feet of the perpendiculars drawn from the vertex of the parabola $y^2 = 4ax$ upon all such chords of the parabola which subtend a right angle at the vertex is

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