The normal to the parabola {y^2 = 8x} at the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of the parabola is ${y^2 = 8x}$,which is of the form ${y^2 = 4ax}$,so ${a = 2}$.The point $(2, 4)$ corresponds to the parameter ${t_1

Vedclass · Accepted Answer

$\{18, -12\}$

Vedclass · Answer

(D) The equation of the parabola is ${y^2 = 8x}$,which is of the form ${y^2 = 4ax}$,so ${a = 2}$.The point $(2, 4)$ corresponds to the parameter ${t_1}$ where ${at_1^2 = 2}$ and ${2at_1 = 4}$.Substituting ${a = 2}$,we get ${2t_1^2 = 2 \implies t_1^2 = 1 \implies t_1 = 1}$ (since ${y > 0}$).If the normal at ${t_1}$ meets the parabola at ${t_2}$,then ${t_2 = -t_1 - \frac{2}{t_1}}$.Substituting ${t_1 = 1}$,we get ${t_2 = -1 - \frac{2}{1} = -3}$.The coordinates of the point are $(at_2^2, 2at_2) = (2(-3)^2, 2(2)(-3)) = (18, -12)$.

The normal to the parabola ${y^2 = 8x}$ at the point $(2, 4)$ meets the parabola again at the point

Similar Questions

Statement $- 1 :$ For all non-zero values of $m$,$y = mx - 1/m$ is always a tangent to the parabola $y^2 = -4x$.
Statement $- 2 :$ Every tangent to the parabola $y^2 = -4x$ touches its axis at a point whose $x$-coordinate is non-negative.

If $a$ and $c$ are the segments of a focal chord of a parabola and $b$ is the semi-latus rectum,then:

What is the slope of the tangent drawn from the point $(4, 10)$ to the parabola $y^2 = 9x$?

If the line $y = 2x + k$ is a normal to the parabola $y^2 = 4x$ at the point $(t^2, 2t)$,then:

Find the equation of the tangent to the curve $y^2 = 6x$ at the point $(2, -3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module