JEE Main 2024 — 1 February, Shift 1 — Question 18

$\mathrm{x}^{2}+\mathrm{y}^{2}=4$

$\mathrm{C}(0,0)$

$\mathrm{r}_{1}=2$

$\mathrm{C}^{\prime}(2 \lambda, 0)$

$\mathrm{r}_{2}=\sqrt{4 \lambda^{2}-9}$

$\left|\mathrm{r}_{1}\mathrm{r}_{2}\right|<\mathrm{CC}^{\prime}<\left|\mathrm{r}_{1}+\mathrm{r}_{2}\right|$

$\left|2-\sqrt{4 \lambda^{2}-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^{2}-9}$

$4+4 \lambda^{2}-9-4 \sqrt{4 \lambda^{2}-9}<4 \lambda^{2}$

True $\lambda \in \operatorname{R} \ldots . .(1)$

$4 \lambda^{2}<4+4 \lambda^{2}-9+4 \sqrt{4 \lambda^{2}-9}$

$5<4 \sqrt{4 \lambda^{2}-9}$ and $\quad \lambda^{2} \geq \frac{9}{4}$

$\frac{25}{16}<4 \lambda^{2}-9 \quad \lambda \in\left(-\infty,-\frac{3}{2}\right] \cup\left[\frac{3}{2}, \infty\right)$

$\frac{169}{64}<\lambda^{2}$

$\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right)$

from (1) and (2)

$\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \Rightarrow \mathrm{R}-\left[-\frac{13}{8}, \frac{13}{8}\right]$

as per question $\mathrm{a}=-\frac{13}{8}$ and $\mathrm{b}=\frac{13}{8}$

$\therefore \quad$ required point is $(-1,6)$ with satisfies option (4)

$6 x^{2}+y^{2}=42$