JEE Mathematics 2024 Question with Solution

Given: The parabola has vertex $\left(2, 3\right)$ and directrix $2x + y = 6$. The ellipse has eccentricity $\frac{1}{\sqrt{2}}$ and passes through the focus of the parabola.

Find: The square of the latus rectum of the ellipse.

The distance from the vertex to the directrix is

$$\left| \frac{2 \cdot 2 + 1 \cdot 3 - 6}{\sqrt{2^2 + 1^2}} \right| = \frac{1}{\sqrt{5}}$$

So the focus of the parabola is

$$\left( 2 + \frac{2}{\sqrt{5}}, 3 + \frac{1}{\sqrt{5}} \right) = \left( 2.4, 3.2 \right)$$

For the ellipse,

$$e = \frac{\sqrt{a^2 - b^2}}{a} = \frac{1}{\sqrt{2}}$$

Hence,

$$1 - \frac{b^2}{a^2} = \frac{1}{2}$$

which gives

$$a^2 = 2b^2$$

Since the ellipse passes through $\left(2.4, 3.2\right)$,

$$\left( \frac{24}{10} \right)^2 \frac{1}{a^2} + \left( \frac{32}{10} \right)^2 \frac{1}{b^2} = 1$$

Substituting $a^2 = 2b^2$,

$$\left( \frac{24}{10} \right)^2 \frac{1}{2b^2} + \left( \frac{32}{10} \right)^2 \frac{1}{b^2} = 1$$

This gives

$$b^2 = \frac{328}{25}$$

Now the square of the latus rectum is taken in the solution as

$$\left( \frac{2b^2}{a} \right)^2 = \frac{4b^4}{a^2}$$

Using $a^2 = 2b^2$,

$$\left( \frac{2b^2}{a} \right)^2 = 4 \times \frac{1}{2} \times \frac{328}{25} = \frac{656}{25}$$

Therefore, the correct option is D.

Given: Vertex of parabola $\left(2, 3\right)$, directrix $2x + y = 6$, and eccentricity of ellipse $\frac{1}{\sqrt{2}}$.

Find: The square of the latus rectum of the ellipse.

The directrix is

$$2x + y - 6 = 0$$

Its normal direction is $\left(2, 1\right)$, so the axis of the parabola is along this direction. The perpendicular distance from the vertex $\left(2, 3\right)$ to the directrix is

$$\frac{|2 \cdot 2 + 1 \cdot 3 - 6|}{\sqrt{2^2 + 1^2}} = \frac{1}{\sqrt{5}}$$

Hence the focus lies at the same distance from the vertex along the normal direction:

$$\left(2, 3\right) + \frac{1}{\sqrt{5}} \cdot \left( \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right) = \left( 2 + \frac{2}{\sqrt{5}}, 3 + \frac{1}{\sqrt{5}} \right)$$

The solution writes this point as

$$\left(2.4, 3.2\right)$$

Now for the ellipse,

$$e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$$

so

$$\frac{b^2}{a^2} = \frac{1}{2}$$

and therefore

$$a^2 = 2b^2$$

Using the point condition from the focus of the parabola,

$$\frac{\left(24/10\right)^2}{a^2} + \frac{\left(32/10\right)^2}{b^2} = 1$$

Replacing $a^2$ by $2b^2$,

$$\frac{\left(24/10\right)^2}{2b^2} + \frac{\left(32/10\right)^2}{b^2} = 1$$

which yields

$$b^2 = \frac{328}{25}$$

Then

$$\left( \frac{2b^2}{a} \right)^2 = \frac{4b^4}{a^2} = \frac{4b^4}{2b^2} = 2b^2 = 2 \cdot \frac{328}{25} = \frac{656}{25}$$

Therefore, the square of the latus rectum is $\frac{656}{25}$, so the correct option is D.