JEE Mathematics 2026 Question with Solution

Given: The ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $P(\alpha,\beta)$ lies on it in the first quadrant.

Find: The value of $\alpha^2 + \beta^2$ given that

$$(SP)^2 + (S'P)^2 - SP \cdot S'P = 37$$

For the ellipse,

$$a^2 = 25, \quad b^2 = 9$$

so

$$c^2 = a^2 - b^2 = 25 - 9 = 16 \Rightarrow c = 4$$

Hence the foci are $S(4,0)$ and $S'(-4,0)$.

For any point on the ellipse,

$$SP + S'P = 2a = 10$$

Now use the identity

$$(SP)^2 + (S'P)^2 - SP \cdot S'P = (SP + S'P)^2 - 3(SP)(S'P)$$

Substituting the given value,

$$100 - 3(SP)(S'P) = 37$$

So,

$$(SP)(S'P) = 21$$

Next,

$$SP = \sqrt{(\alpha-4)^2 + \beta^2}, \quad S'P = \sqrt{(\alpha+4)^2 + \beta^2}$$

Therefore,

$$SP \cdot S'P = \sqrt{(\alpha-4)^2 + \beta^2}\sqrt{(\alpha+4)^2 + \beta^2}$$

Squaring,

$$(SP \cdot S'P)^2 = \left[(\alpha^2+\beta^2+16)^2 - (8\alpha)^2\right]$$

Since $SP \cdot S'P = 21$,

$$(\alpha^2+\beta^2+16)^2 - 64\alpha^2 = 441$$

Also, from the ellipse equation,

$$\frac{\alpha^2}{25} + \frac{\beta^2}{9} = 1$$

which gives

$$9\alpha^2 + 25\beta^2 = 225$$

Solving these simultaneously gives

$$\alpha^2 + \beta^2 = 13$$

Therefore, the correct option is B.

Given: $\frac{x^2}{25} + \frac{y^2}{9} = 1$ with foci $S, S'$ and point $P(\alpha,\beta)$ on the ellipse.

Find: $\alpha^2 + \beta^2$.

First identify the ellipse parameters:

$$a = 5, \quad b = 3, \quad c = 4$$

so the foci are $S(4,0)$ and $S'(-4,0)$.

Since $P$ lies on the ellipse,

$$SP + S'P = 2a = 10$$

The given condition is

$$(SP)^2 + (S'P)^2 - SP \cdot S'P = 37$$

Rewrite the first two terms using

$$(SP + S'P)^2 = (SP)^2 + (S'P)^2 + 2(SP)(S'P)$$

Hence,

$$(SP)^2 + (S'P)^2 - SP \cdot S'P = (SP + S'P)^2 - 3(SP)(S'P)$$

Substitute $SP + S'P = 10$:

$$10^2 - 3(SP)(S'P) = 37$$ $$100 - 3(SP)(S'P) = 37$$ $$(SP)(S'P) = 21$$

Now write the distances from the foci:

$$SP = \sqrt{(\alpha-4)^2 + \beta^2}$$ $$S'P = \sqrt{(\alpha+4)^2 + \beta^2}$$

Thus,

$$SP \cdot S'P = \sqrt{\big((\alpha-4)^2 + \beta^2\big)\big((\alpha+4)^2 + \beta^2\big)}$$

Squaring both sides,

$$441 = \big((\alpha-4)^2 + \beta^2\big)\big((\alpha+4)^2 + \beta^2\big)$$

Expanding in the form used in the solution,

$$441 = (\alpha^2 + \beta^2 + 16)^2 - 64\alpha^2$$

Also, because $P(\alpha,\beta)$ lies on the ellipse,

$$\frac{\alpha^2}{25} + \frac{\beta^2}{9} = 1$$ $$9\alpha^2 + 25\beta^2 = 225$$

Using these two equations, we obtain

$$\alpha^2 + \beta^2 = 13$$

Therefore, the correct option is B.