If P is a point on the parabola y^2=8x and A | vedclass

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

Question

(A) Let $P$ be a point on the parabola $y^2=8x$. The coordinates of $P$ can be represented as $(2t^2, 4t)$.Let $M(x, y)$ be the mid-point of the line

Vedclass · Accepted Answer

$y^2=4(x-\frac{1}{2})$

Vedclass · Answer

(A) Let $P$ be a point on the parabola $y^2=8x$. The coordinates of $P$ can be represented as $(2t^2, 4t)$.Let $M(x, y)$ be the mid-point of the line segment $AP$,where $A$ is $(1, 0)$.Using the mid-point formula,we have:$x = \frac{2t^2 + 1}{2}$ and $y = \frac{4t + 0}{2} = 2t$.From $y = 2t$,we get $t = \frac{y}{2}$.Substituting $t$ into the equation for $x$:$x = \frac{2(\frac{y}{2})^2 + 1}{2} = \frac{2(\frac{y^2}{4}) + 1}{2} = \frac{\frac{y^2}{2} + 1}{2} = \frac{y^2 + 2}{4}$.$4x = y^2 + 2$.$y^2 = 4x - 2 = 4(x - \frac{1}{2})$.Thus,the locus of the mid-point is $y^2 = 4(x - \frac{1}{2})$.

If $P$ is a point on the parabola $y^2=8x$ and $A$ is the point $(1,0)$,then the locus of the mid-point of the line segment $AP$ is

Similar Questions

The tangent at the point $(1, 2)$ to the curve $y^2 = 4x$ makes an angle $\theta$ with the positive direction of the $X$-axis. Then $\theta =$ (in $^{\circ}$)

The equation of the tangent at $(-4, -4)$ on the curve $x^2 = -4y$ is

The equation of the parabola with focus $(3, 0)$ and the directrix $x + 3 = 0$ is

If $P$ is $(3, 1)$ and $Q$ is a point on the curve $y^2 = 8x$,then the locus of the mid-point of the line segment $PQ$ is

Let the tangent and normal at any point $P(at^2, 2at)$,$(a > 0)$,on the parabola $y^2 = 4ax$ meet the axis of the parabola at $T$ and $G$ respectively. Then the radius of the circle through $P, T$ and $G$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module