The straight line joining any point P on the | vedclass

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

Question

(B) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$ $....(1)$The focus of the par

Vedclass · Accepted Answer

$2x^2 + y^2 - 2ax = 0$

Vedclass · Answer

(B) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$ $....(1)$The focus of the parabola is $S(a, 0)$.The line perpendicular to the tangent $(1)$ passing through $S(a, 0)$ has the slope $-t$. Its equation is $y - 0 = -t(x - a)$,which simplifies to $tx + y = at$ $....(2)$The line joining the vertex $O(0, 0)$ to $P(at^2, 2at)$ has the slope $\frac{2at}{at^2} = \frac{2}{t}$. Its equation is $y = \frac{2}{t}x$,or $2x - ty = 0$ $....(3)$To find the locus of $R(h, k)$,we eliminate $t$ from $(2)$ and $(3)$.From $(3)$,$t = \frac{2h}{k}$.Substituting this into $(2)$: $(\frac{2h}{k})h + k = a(\frac{2h}{k})$.Multiplying by $k$: $2h^2 + k^2 = 2ah$.Replacing $(h, k)$ with $(x, y)$,the locus is $2x^2 + y^2 - 2ax = 0$.

The straight line joining any point $P$ on the parabola $y^2 = 4ax$ to the vertex and the perpendicular from the focus to the tangent at $P$ intersect at $R$. Then the equation of the locus of $R$ is:

Similar Questions

Find the point where the normal drawn from the upper end of the latus rectum of the parabola $y^2 = -12x$ intersects the axis.

The point on the parabola $2y = x^2$ which is nearest to the point $(0, 3)$ is

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the slope of the tangent drawn at $(1, k)$ on the given parabola is

The equation of the parabola whose directrix is $y = 2x - 9$ and focus is $(-8, -2)$ is:

$A$ tangent to the curve $x = a t^{2}, y = 2 a t$ is perpendicular to the $X$-axis. Then the point of contact is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module