The normal drawn at a point (2, -4) on the pa | vedclass

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

Question

(D) Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.For a point $(at^2, 2at)$ on the parabola,we have $(2, -4) = (2t^2, 4t)$,which gives $t

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.For a point $(at^2, 2at)$ on the parabola,we have $(2, -4) = (2t^2, 4t)$,which gives $t = -1$.The normal at parameter $t$ meets the parabola again at parameter $t_2 = -t - \frac{2}{t}$.Substituting $t = -1$,we get $t_2 = -(-1) - \frac{2}{-1} = 1 + 2 = 3$.The coordinates of the point $(\alpha, \beta)$ are $(at_2^2, 2at_2) = (2(3)^2, 2(2)(3)) = (18, 12)$.Thus,$\alpha + \beta = 18 + 12 = 30$.

The normal drawn at a point $(2, -4)$ on the parabola $y^2 = 8x$ cuts the same parabola again at $(\alpha, \beta)$. Then $\alpha + \beta =$

Similar Questions

Find the length of the chord of the parabola $y^2 = 4ax$ which passes through the vertex and makes an angle $\theta$ with the $x-$axis.

$(1, 1)$ is the vertex and $x+y+1=0$ is the directrix of a parabola. If $(a, b)$ is its focus and $(c, d)$ is the point of intersection of the directrix and the axis of the parabola,then $a+b+c+d=$

One of the points on the parabola $y^2 = 12x$ with focal distance $12$ is:

If the focus of a parabola is $(3, -4)$ and its directrix is $x + y - 2 = 0$,find the vertex.

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 11x$,then $2(m_1^2 + m_2^2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module