JEE Main 2024 — 27 January, Shift 1 — Question 82

Equation of chord with given middle point: $T = S_1$

$\frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{1}{100}$

$\frac{8x + 5y}{200} = \frac{8 + 2}{200}$

$y = \frac{10 - 8x}{5}$ ...(i)

$\frac{x^2}{25} + \frac{(10 - 8x)^2}{400} = 1$ (put in original equation)

$\frac{16x^2 + 100 + 64x^2 - 160x}{400} = 1$

$4x^2 - 8x - 15 = 0$

$x = \frac{8 \pm \sqrt{304}}{8}$

$x_1 = \frac{8 + \sqrt{304}}{8} \text{; } x_2 = \frac{8 - \sqrt{304}}{8}$

Similarly, $y = \frac{10 - 18 \pm \sqrt{304}}{5} = \frac{2 \pm \sqrt{304}}{5}$

$y_1 = \frac{2 - \sqrt{304}}{5} \text{; } y_2 = \frac{2 + \sqrt{304}}{5}$

$\text{Distance} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$= \sqrt{\frac{4 \times 304}{64} + \frac{4 \times 304}{25}} = \frac{\sqrt{1691}}{5}$