Question

If the point $P(\alpha ,-\alpha )$ lies inside the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1,$ then

• A. $\alpha \in (-\infty ,-\frac{12}{5})\cup (\frac{12}{5},\infty )$
• B. $\alpha \in (-\frac{12}{5},\frac{12}{5})$
• C. $\alpha \in (-\frac{5}{12},\frac{5}{12})$
• D. $\alpha \in (-\infty ,-\frac{5}{12})\cup (\frac{5}{12},\infty )$

Answer 1

The correct option is B $\alpha \in (-\frac{12}{5},\frac{12}{5})$

If $P(\alpha ,-\alpha )$ lies inside the ellipse, then
$S_1<0\Rightarrow \frac{{\alpha }^2}{16}+\frac{{\alpha }^2}{9}-1<0$
$\Rightarrow \frac{25{\alpha }^2}{144}<1$
$\Rightarrow {\alpha }^2<\frac{144}{25}\therefore \alpha \in (-\frac{12}{5},\frac{12}{5})$