The centroid of the triangle formed by the feet of the normals from the point (h, k) to the parabola y2+4 a x =0,(a0) lies on

Question

The centroid of the triangle formed by the feet of the normals from the point (h, k) to the parabola y2+4 a x =0,(a>0) lies on (A) x-axis (B) y-axis (C) x=h (D) y=k

Tardigrade Admin · Accepted Answer

Co-ordinates of any point on the parabola

y^{2} = - 4 a x

are

(- a t^{2}, 2 a t)

Equation of the normal at

(- a t^{2}, 2 a t)

is

y - x t = 2 a t + a t^{3}

If the normal passes through the point

(h, k)

, then
or

a t^{3} + (2 a + h) t - k = 0

,

k - t h = 2 a t + a t^{3}

which is a cubic equation whose three roots

t_{1}, t_{2}, t_{3}

are the parameters of the feet of the three normals.

∴

Sum of the roots

= t_{1} + t_{2} + t_{3} = - \frac{Coefficient of t ^{2}}{Coefficient of t ^{3}} = 0

∴

Centroid of the triangle formed by the feet of the normals

= (- \frac{a}{3} (t_{1}^{2} + t_{2}^{2} + t_{3}^{2}), \frac{2 a}{3} (t_{1} + t_{2} + t_{3}))

= (- \frac{a}{3} (t_{1}^{2} + t_{2}^{2} + t_{3}^{2}), 0)

which, clearly, lies on the

x

-axis.

Q. The centroid of the triangle formed by the feet of the normals from the point $(h, k)$ to the parabola $y^{2} + 4 a x$ $= 0, (a > 0)$ lies on

$x$ -axis

$y$ -axis

$x = h$

$y = k$

Q. The centroid of the triangle formed by the feet of the normals from the point (h,k) to the parabola y2+4ax =0,(a>0) lies on

x-axis

y-axis

x=h

y=k

Q. The centroid of the triangle formed by the feet of the normals from the point $(h, k)$ to the parabola $y^{2} + 4 a x$ $= 0, (a > 0)$ lies on

$x$ -axis

$y$ -axis

$x = h$

$y = k$