Let P be the point (1, 0) and Q be a point on | vedclass

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

Question

(A) Let the coordinates of point $Q$ on the parabola $y^2 = 8x$ be $(2t^2, 4t)$.Let the midpoint of $PQ$ be $(h, k)$.Since $P = (1, 0)$ and $Q = (2t^2

Vedclass · Accepted Answer

$y^2 - 4x + 2 = 0$

Vedclass · Answer

(A) Let the coordinates of point $Q$ on the parabola $y^2 = 8x$ be $(2t^2, 4t)$.Let the midpoint of $PQ$ be $(h, k)$.Since $P = (1, 0)$ and $Q = (2t^2, 4t)$,the midpoint $(h, k)$ is given by:$h = \frac{2t^2 + 1}{2} \implies 2t^2 = 2h - 1$$k = \frac{4t + 0}{2} = 2t \implies t = \frac{k}{2}$Substitute $t = \frac{k}{2}$ into the equation $2t^2 = 2h - 1$:$2(\frac{k}{2})^2 = 2h - 1$$2(\frac{k^2}{4}) = 2h - 1$$\frac{k^2}{2} = 2h - 1$$k^2 = 4h - 2$Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 4x - 2$,which is $y^2 - 4x + 2 = 0$.

Let $P$ be the point $(1, 0)$ and $Q$ be a point on the parabola $y^2 = 8x$. Find the locus of the midpoint of $PQ$.

Similar Questions

From a point $(C, 0)$,three normals are drawn to the parabola $y^2=x$. Then,

If $P\left(\frac{1}{2}, 4\right)$ and $Q$ are the ends of a focal chord of the parabola $y^2=32x$ and $S$ is the focus of the parabola,then $SQ=$

If the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again at $Q$,then the coordinates of $Q$ are

The length of the latus rectum of the parabola $4y^2 + 2x - 20y + 17 = 0$ is

The equation of the tangent to the parabola $y = x^2 - x$ at the point where $x = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module