Question

Find the set of values of $\alpha$ for which point the $P(\alpha ,-\alpha )$ is inside
$\frac{x^2}{16}+\frac{y^2}{9}=1$

• A. $(\frac{-12}{5},\frac{12}{5})$
• B. $[\frac{-12}{5},\frac{12}{5}]$
• C. $(\frac{-13}{5},\frac{13}{5})$
• D. $(\frac{-13}{5},\frac{13}{5}]$

Answer 1

The correct option is A $(\frac{-12}{5},\frac{12}{5})$

Let $P(x_1,y_1)$ be the point, where $x_1=\alpha ,y_1=-\alpha$.
$\therefore \text{If}{S}_1=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$ $<0$ then P is inside.
$\therefore S_1=\frac{{\alpha }^2}{16}+\frac{{(-\alpha )}^2}{9}-1$
$=\frac{25{\left(\alpha \right)}^2}{16.9}-1$$<0$
$\Rightarrow \frac{25{\alpha }^2}{16.9}$ $<1$ $\Rightarrow {\alpha }^2$ $<$ $\frac{16.9}{25}$
$\therefore \left|\alpha \right|$ $<$ $\frac{4.3}{5}=\frac{12}{5}$
$\Rightarrow \alpha \in (\frac{-12}{5},\frac{12}{5})$.