JEE Main 2025 — 28 January, Evening Shift — Question 19

$x^{2}+y^{2}-8 x=0, \frac{x^{2}}{9}-\frac{y^{2}}{4}=1$

$4 x^{2}-9 y^{2}=36$

Solve (1) & (2)

$4 x^{2}-9\left(8 x-x^{2}\right)=36$

$13 x^{2}-72 x-36=0$

$(13 \mathrm{x}+6)(\mathrm{x}-6)=0$

$x=\frac{-6}{13}, x=6$

$\mathrm{x}=\frac{-6}{13}($ rejected $)$

$\mathrm{y} \rightarrow$ Imaginary

$\mathrm{x}=6, \frac{36}{9}-\frac{\mathrm{y}^{2}}{4}=1$

$\mathrm{y}^{2}=12, \mathrm{y}=\mathrm{+ or-} \sqrt{12}$

$\mathrm{A}(6, \sqrt{12}), \mathrm{B}(6,-\sqrt{12})$

$\mathrm{p}\left(\alpha, \frac{2 \alpha+4}{3}\right) \mathrm{P}$ lies on 2 x-3 y+4=0$

$\mathrm{h}=\frac{12+\alpha}{3}, \alpha=3 \mathrm{~h}-12$

$\mathrm{k}=\frac{\frac{2 \alpha+4}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{k}$

$\alpha=\frac{9 \mathrm{k}-4}{2}$

$6 \mathrm{~h}-24=9 \mathrm{k}-4$

$6 x-9 y=20$