JEE Mathematics 2026 Question with Solution

Given: The ellipse is $3x^2 + 4y^2 = 1$ and the tangent is $ax + 4y = \sqrt{7}$ touching it at point $P(x_1,y_1)$ in the first quadrant.

Find: One focal distance of the point $P$.

Write the ellipse as

$$\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1$$

So, for the standard ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, we have

$$A^2 = \frac{1}{3}, \qquad B^2 = \frac{1}{4}$$

Hence

$$A = \frac{1}{\sqrt{3}}, \qquad B = \frac{1}{2}$$

The tangent at $P(x_1,y_1)$ to the ellipse $3x^2 + 4y^2 = 1$ is

$$3xx_1 + 4yy_1 = 1$$

The given tangent can be written as

$$\frac{a}{\sqrt{7}}x + \frac{4}{\sqrt{7}}y = 1$$

Comparing coefficients with the tangent form,

$$4y_1 = \frac{4}{\sqrt{7}} \implies y_1 = \frac{1}{\sqrt{7}}$$

Since $P$ lies on the ellipse,

$$3x_1^2 + 4y_1^2 = 1$$

Substituting $y_1 = \frac{1}{\sqrt{7}}$,

$$3x_1^2 + 4\left(\frac{1}{7}\right) = 1$$ $$3x_1^2 = 1 - \frac{4}{7} = \frac{3}{7}$$ $$x_1^2 = \frac{1}{7}$$

Since $P$ is in the first quadrant,

$$x_1 = \frac{1}{\sqrt{7}}$$

Now the eccentricity is

$$e^2 = 1 - \frac{B^2}{A^2} = 1 - \frac{1/4}{1/3} = 1 - \frac{3}{4} = \frac{1}{4}$$

So

$$e = \frac{1}{2}$$

For a point $P(x_1,y_1)$ on the ellipse, the focal distances are

$$A \pm ex_1$$

Therefore,

$$\text{focal distances} = \frac{1}{\sqrt{3}} \pm \frac{1}{2}\cdot\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}$$

Hence one focal distance is $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$. Therefore, the correct option is B.

Given: The ellipse is $3x^2 + 4y^2 = 1$ and the point of contact $P$ lies in the first quadrant.

Find: One focal distance of $P$.

For the ellipse

$$\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1$$

the semi-major axis is

$$a = \frac{1}{\sqrt{3}}$$

and eccentricity is

$$e = \frac{1}{2}$$

From the tangent comparison as above, the point of contact is

$$P\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$$

For an ellipse, the distances of $P(x_1,y_1)$ from the two foci are

$$a \pm ex_1$$

Substituting $a = \frac{1}{\sqrt{3}}, e = \frac{1}{2}, x_1 = \frac{1}{\sqrt{7}}$,

$$\frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}$$

Also, their sum is

$$2a = \frac{2}{\sqrt{3}}$$

which is consistent with the ellipse property. Thus one focal distance is $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$, so the correct option is B.