JEE Mathematics 2025 Question with Solution

Given: Focus is $(-2, 1)$ and directrix is $2x + y + 2 = 0$.

Find: The sum of the ordinates of the points on the parabola whose abscissa is $-2$.

For any point $P(x, y)$ on a parabola, the distance from the point to the focus equals the perpendicular distance from the point to the directrix.

So,

$$\sqrt{(x+2)^2 + (y-1)^2} = \frac{|2x + y + 2|}{\sqrt{5}}$$

Squaring both sides,

$$(x+2)^2 + (y-1)^2 = \left(\frac{2x + y + 2}{\sqrt{5}}\right)^2$$ $$5\left[(x+2)^2 + (y-1)^2\right] = (2x + y + 2)^2$$

Now substitute $x = -2$:

$$5(y-1)^2 = (y-2)^2$$

Expanding,

$$5(y^2 - 2y + 1) = y^2 - 4y + 4$$ $$5y^2 - 10y + 5 = y^2 - 4y + 4$$ $$4y^2 - 6y + 1 = 0$$

If the two ordinates are $y_1$ and $y_2$, then by sum of roots,

$$y_1 + y_2 = \frac{-(-6)}{4} = \frac{6}{4} = \frac{3}{2}$$

Therefore, the sum of the ordinates is $\frac{3}{2}$. The correct option is A.

Given: Focus is $F(-2, 1)$ and directrix is $2x + y + 2 = 0$.

Find: The sum of the ordinates of the points on the parabola with abscissa $-2$.

Take a point on the parabola as $(-2, y)$.

Distance from $(-2, y)$ to the focus $(-2, 1)$ is

$$\sqrt{(-2+2)^2 + (y-1)^2} = \sqrt{(y-1)^2}$$

Distance from $(-2, y)$ to the directrix $2x + y + 2 = 0$ is

$$\frac{|2(-2) + y + 2|}{\sqrt{2^2 + 1^2}} = \frac{|y-2|}{\sqrt{5}}$$

Using the parabola definition,

$$\sqrt{(y-1)^2} = \frac{|y-2|}{\sqrt{5}}$$

Squaring,

$$(y-1)^2 = \frac{(y-2)^2}{5}$$ $$5(y-1)^2 = (y-2)^2$$

This gives

$$4y^2 - 6y + 1 = 0$$

Hence the sum of the two possible ordinates is

$$\frac{6}{4} = \frac{3}{2}$$

Therefore, the required sum is $\frac{3}{2}$.