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

Question

The area bounded by $ y - 1 = |x| $ and $ y + 1 = |x| $ is: (a) $ \frac{1}{2} $

Answer 1

To determine the locus of the point $P$, which divides the line segment $OQ$ internally in the ratio $1 : 3$, we start by identifying the coordinates of points $O$ and $Q$.

Here, the vertex of the parabola $x^2 = 8y$ is at the origin $O(0, 0)$. Let $Q(x_1, y_1)$ be any point on the parabola.

Since $Q$ lies on the parabola, we have $y_1 = \frac{x_1^2}{8}$.

According to the section formula, if point $P$ divides the segment $OQ$ in the ratio $1:3$, the coordinates of $P$ will be given by:

$P\left(\frac{1 \cdot x_1 + 3 \cdot 0}{1 + 3}, \frac{1 \cdot y_1 + 3 \cdot 0}{1 + 3}\right) = \left(\frac{x_1}{4}, \frac{y_1}{4}\right)$.

Substituting $y_1 = \frac{x_1^2}{8}$ into the coordinates of $P$, we have:

$P\left(\frac{x_1}{4}, \frac{\frac{x_1^2}{8}}{4}\right) = \left(\frac{x_1}{4}, \frac{x_1^2}{32}\right)$.

Let the coordinates of point $P$ be $(x, y)$. Then, we have:

$x = \frac{x_1}{4}$
$y = \frac{x_1^2}{32}$

From $x = \frac{x_1}{4}$, we get $x_1 = 4x$.

Substituting $x_1 = 4x$ into the equation for $y$, we get:

$y = \frac{(4x)^2}{32} = \frac{16x^2}{32} = \frac{x^2}{2}$.

Thus, the locus of $P$ is given by $x^2 = 2y$.

Therefore, the correct answer 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

The Correct Option is D

Solution and Explanation

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