Let $PQ$ be a double ordinate of the parabola, $y^2= - 4x$, where P lies in the second quadrant. If R divides $PQ$ in the ratio $2 : 1$, then the locus of R is :

Question

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

Answer 1

To solve this problem, we need to find the locus of the point R which divides the double ordinate PQ of the parabola $y^2 = -4x$ in the ratio $2:1$.

First, understand that the parabola $y^2 = -4x$ is downward facing and symmetric about the x-axis. The double ordinate PQ means both points P and Q have the same x-coordinate but different y-coordinates.
Let's assume the coordinates of P and Q are $(x_1, y_1)$ and $(x_1, y_2)$, respectively. Since PQ is a double ordinate, they satisfy $y_1^2 = -4x_1$ and $y_2^2 = -4x_1$.
R divides PQ in the ratio $2:1$. Therefore, the coordinates of R can be calculated using the section formula. Let's denote R by $(x, y)$. Using the section formula, the coordinates are given by: \[ x = x_1 \] \[ y = \frac{2y_2 + y_1}{3} \]
For P to lie in the second quadrant, $x_1$ should be negative, and $y_1$ will be positive. Due to the symmetry, $y_2 = -y_1$.
Substitute $y_2 = -y_1$: \[ y = \frac{2(-y_1) + y_1}{3} = \frac{-2y_1 + y_1}{3} = -\frac{y_1}{3} \]
From $y_1^2 = -4x_1$, we have $x_1 = -\frac{y_1^2}{4}$. Substitute into $y^2$: \[ y^2 = \left(-\frac{y_1}{3}\right)^2 = \frac{y_1^2}{9} \]
Now substitute for $x_1$ to eliminate $y_1$: \[ x = -\frac{y_1^2}{4} \] Equating the expression for $y^2$: \[ 9y^2 = y_1^2 \] Substitute in $x_1 = -\frac{y_1^2}{4}$: \[ x = -\frac{9y^2}{4} \]
The relationship between $x$ and $y$ gives the locus of point R as: \[ 9y^2 = -4x \]

Therefore, the locus of R is $9y^2 = -4x$, which matches the provided correct answer.

The Correct Option is B