Let O be the vertex and Q be any point on the | vedclass

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

Question

(D) The vertex of the parabola $x^2=8y$ is $O(0, 0)$.Let the coordinates of point $Q$ be $(x_1, y_1)$. Since $Q$ lies on the parabola, $x_1^2 = 8y_1$.

Vedclass · Accepted Answer

$x^2=2y$

Vedclass · Answer

(D) The vertex of the parabola $x^2=8y$ is $O(0, 0)$.Let the coordinates of point $Q$ be $(x_1, y_1)$. Since $Q$ lies on the parabola, $x_1^2 = 8y_1$.Let the coordinates of point $P$ be $(h, k)$.Since $P$ divides $OQ$ in the ratio $1:3$, by the section formula:$h = \frac{1 \cdot x_1 + 3 \cdot 0}{1+3} = \frac{x_1}{4} \Rightarrow x_1 = 4h$$k = \frac{1 \cdot y_1 + 3 \cdot 0}{1+3} = \frac{y_1}{4} \Rightarrow y_1 = 4k$Substituting these into the parabola equation $x_1^2 = 8y_1$:$(4h)^2 = 8(4k)$$16h^2 = 32k$$h^2 = 2k$Replacing $(h, k)$ with $(x, y)$, the locus of $P$ is $x^2 = 2y$.

Let $O$ be the vertex and $Q$ be any point on the parabola $x^2=8y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1:3$, then the locus of $P$ is

Similar Questions

The line $y = mx + c$ touches the parabola $x^2 = 4ay$ if:

If the line $x + y = 1$ is a normal to the parabola $y^2 = kx$,then find the value of $k$. (in $/3$)

The equation of the chord of contact of the tangents drawn from the point $(2, 3)$ to the parabola $y^2 + x = 0$ is:

If the normal at the point $t_1$ (i.e.,at $(at_1^2, 2at_1)$) on the parabola $y^2 = 4ax$ meets the parabola again at the point $t_2$,then $t_1t_2$ is equal to:

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module