Let the locus of the mid-point of the chord t | vedclass

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

Question

(B) Let the parabola be $y^2=4x$. Any point on the parabola is $P(t^2, 2t)$.The chord passes through the origin $O(0,0)$ and $P(t^2, 2t)$.The mid-poin

Vedclass · Accepted Answer

$2y^2=3x$

Vedclass · Answer

(B) Let the parabola be $y^2=4x$. Any point on the parabola is $P(t^2, 2t)$.The chord passes through the origin $O(0,0)$ and $P(t^2, 2t)$.The mid-point $M(h, k)$ of the chord $OP$ is given by $h = \frac{t^2+0}{2} = \frac{t^2}{2}$ and $k = \frac{2t+0}{2} = t$.Substituting $t=k$ into $h = \frac{t^2}{2}$, we get $h = \frac{k^2}{2}$, so $k^2=2h$.The locus $S$ is $y^2=2x$.Let $P(x_0, y_0)$ be a point on $S$, so $y_0^2=2x_0$. Since $P$ is on $S$, we can write $P$ as $(2t^2, 2t)$ because $(2t)^2 = 2(2t^2)$.Let $R(x, y)$ be the point that divides $OP$ internally in the ratio $3:1$. Using the section formula:$x = \frac{3(2t^2) + 1(0)}{3+1} = \frac{6t^2}{4} = \frac{3t^2}{2}$$y = \frac{3(2t) + 1(0)}{3+1} = \frac{6t}{4} = \frac{3t}{2}$From $y = \frac{3t}{2}$, we get $t = \frac{2y}{3}$.Substituting $t$ into $x = \frac{3t^2}{2}$:$x = \frac{3}{2} \left(\frac{2y}{3}\right)^2 = \frac{3}{2} \cdot \frac{4y^2}{9} = \frac{2y^2}{3}$Thus, $3x = 2y^2$, or $2y^2=3x$.

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2}=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point, which internally divides $OP$ in the ratio $3:1$, is:

Similar Questions

If $b$ and $c$ are the lengths of the segments of any focal chord of a parabola $y^2 = 4ax$,then the length of the semi-latus rectum is:

The tangent drawn at any point $P$ to the parabola ${y^2} = 4ax$ meets the directrix at the point $K$. Then the angle which $KP$ subtends at its focus is ............. $^\circ$.

If the tangents and normals at the extremities of a focal chord of a parabola $y^2 = 4ax$ intersect at $(x_1, y_1)$ and $(x_2, y_2)$ respectively,then:

The equation of the tangent to the parabola $y^2=12x$,which makes an angle $30^{\circ}$ with the positive direction of the $X$-axis is given by $x-\sqrt{3}y+9=0$. The point of contact is:

$A = (-2, 0)$ and $P$ is a point on the parabola $y^2 = 8x$. If $Q$ bisects $\overline{AP}$ and the locus of $Q$ is a parabola,then its focus is

Vedclass Test Series

Exam Paper Generator

Online Exam Module