Let the normal at the point P on the parabola y 2= 6 x pass through the point (5,-8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :

Question

Let the normal at the point P on the parabola y 2= 6 x pass through the point (5,-8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is : (A) -3 (B) -(9/4) (C) -(5/2) (D) -2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b3bd9900e3955ec1e10eade982d8cd0a-.png" /> Equation of normal: y =- tx +2 at + at 3 ( a =(3/2)) since passing through (5,-8), we get t =-2 Co-ordinate of Q:(6,-6) Equation of tangent at Q: x +2 y +6=0 Put x=(-3/2) to get R ((-3/2), (-9/4))

Q. Let the normal at the point $P$ on the parabola $y^{2} =$ $6 x$ pass through the point $(5, - 8)$ . If the tangent at $P$ to the parabola intersects its directrix at the point $Q$ , then the ordinate of the point $Q$ is :

$- 3$

$- \frac{9}{4}$

$- \frac{5}{2}$

$- 2$

Equation of normal : $y = - t x + 2 a t + a t^{3} (a = \frac{3}{2})$
since passing through $(5, - 8)$ , we get $t = - 2$
Co-ordinate of $Q : (6, - 6)$
Equation of tangent at $Q : x + 2 y + 6 = 0$
Put $x = \frac{- 3}{2}$ to get $R (\frac{- 3}{2}, \frac{- 9}{4})$

Q. Let the normal at the point P on the parabola y2= 6x pass through the point (5,−8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :

−3

−49​

−25​

−2