Question

Locus of point of intersection of normals drawn at end points of focal chord of a parabola $y^2 = 4ax$ is :

• A. y^2 = a(x-a)
• B. y^2 = a(x-2a)
• C. y^2 = a(x-3a)
• D. y^2 = a(x-4a)

Answer 1

Answer: (C)

y^2 = a(x-3a)

Answer 2

The equation of the normal to the parabola $y^2 = 4ax$ at a point $(at^2, 2at)$ is $y = -tx + 2at + at^3$. Let the normals at the endpoints of a focal chord, with parameters $t_1$ and $t_2$, intersect at $(h, k)$. Then $t_1$ and $t_2$ are roots of the cubic equation $at^3 + (2a-h)t + k = 0$. For a focal chord, $t_1t_2 = -1$. Let the third root be $t_3$. From Vieta's formulas, $t_1+t_2+t_3 = 0$, $t_1t_2+t_2t_3+t_3t_1 = \frac{2a-h}{a}$, and $t_1t_2t_3 = -\frac{k}{a}$. Substituting $t_1t_2 = -1$, we get $-t_3 = -\frac{k}{a}$, so $t_3 = \frac{k}{a}$. Also, $-1 + t_3(t_1+t_2) = \frac{2a-h}{a}$. Since $t_1+t_2 = -t_3$, we have $-1 - t_3^2 = \frac{2a-h}{a}$. Substituting $t_3 = \frac{k}{a}$ gives $-1 - (\frac{k}{a})^2 = \frac{2a-h}{a}$. Multiplying by $a$, we get $-a - \frac{k^2}{a} = 2a-h$, which simplifies to $h = 3a + \frac{k^2}{a}$, or $k^2 = a(h-3a)$. Thus, the locus is $y^2 = a(x-3a)$.