Question

Find the set of value(s) of $\alpha$ for which the point $(7-\frac{5}{4}\alpha ,\alpha )$ lies inside the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1.$

• A. $\left(\frac{17}{5}\frac{12}{5}\right)$
• B. $(\frac{12}{5},\frac{16}{5})$
• C. $\frac{-16}{5}$
• D. None of these

Answer 1

Answer: (B)

$(\frac{12}{5},\frac{16}{5})$

$\frac{x^2}{25}+\frac{y^2}{16}=1$

Since point $(7-\frac{5}{4}\alpha ,\alpha )$ lies inside the ellipse

$\therefore S_1<0$

$\Rightarrow 16(7-\frac{5}{4}\alpha {)}^2+25.{\alpha }^2<400$

$\Rightarrow (28-5\alpha {)}^2+25{\alpha }^2<400$

$\Rightarrow 50{\alpha }^2-280\alpha +384<0$

$\Rightarrow 25{\alpha }^2-140\alpha +192<0$

$\Rightarrow (5\alpha -12)(5\alpha -16)<0$

$\Rightarrow \alpha \in (\frac{12}{5},\frac{16}{5})$