JEE Mathematics 2025 Question with Solution

Given: The ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the midpoint of the chord is $\left(\frac{5}{2}, \frac{1}{2}\right)$.

Find: The value of $\alpha + \beta$ if the chord is $\alpha x + \beta y = 109$.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of the chord whose midpoint is $\left(x_1, y_1\right)$ is

$$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$$

Here, $a^2 = 9$, $b^2 = 4$, $x_1 = \frac{5}{2}$ and $y_1 = \frac{1}{2}$.

Substituting these values,

$$\frac{x \cdot (5/2)}{9} + \frac{y \cdot (1/2)}{4} = \frac{(5/2)^2}{9} + \frac{(1/2)^2}{4}$$

So,

$$\frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16}$$

Now simplify the right-hand side:

$$\frac{25}{36} + \frac{1}{16} = \frac{100}{144} + \frac{9}{144} = \frac{109}{144}$$

Therefore,

$$\frac{5x}{18} + \frac{y}{8} = \frac{109}{144}$$

Multiplying throughout by $144$,

$$144\left(\frac{5x}{18} + \frac{y}{8}\right) = 109$$ $$40x + 18y = 109$$

Comparing with $\alpha x + \beta y = 109$, we get

$$\alpha = 40, \quad \beta = 18$$

Hence,

$$\alpha + \beta = 40 + 18 = 58$$

Therefore, the correct option is C.

Given: The chord of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ has midpoint $\left(\frac{5}{2}, \frac{1}{2}\right)$.

Find: The value of $\alpha + \beta$ in $\alpha x + \beta y = 109$.

Using the midpoint form directly,

$$T = S_1$$

For the point $\left(\frac{5}{2}, \frac{1}{2}\right)$,

$$T = \frac{5x}{18} + \frac{y}{8}$$

and

$$S_1 = \frac{(5/2)^2}{9} + \frac{(1/2)^2}{4} = \frac{25}{36} + \frac{1}{16} = \frac{109}{144}$$

Hence the chord is

$$\frac{5x}{18} + \frac{y}{8} = \frac{109}{144}$$

Multiplying by $144$,

$$40x + 18y = 109$$

So $\alpha = 40$ and $\beta = 18$. Therefore,

$$\alpha + \beta = 58$$

Thus, the answer is $58$.