Question

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

Answer 1

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

So: $\frac{x^2}{25}+\frac{y^2}{16}-1$

since $(7-\frac{5\alpha }{4}1\alpha )$ dies inside $S=0$

$\frac{{(7-\frac{5\alpha }{4})}^2}{25}+\frac{{\alpha }^2}{10}+1<0$

$\frac{49+\frac{25{\alpha }^2}{16}-\frac{35\alpha }{2}}{25}+\frac{{\alpha }^2}{16}-1<0$

$\frac{49\times 16+25{\alpha }^2-35\times 8\alpha }{25\times 16}+\frac{{\alpha }^2}{16}-1<0$

$\frac{784+25{\alpha }^2-280\alpha +25{\alpha }^2-16\times 15}{16\times 25}<0$

$784+50{\alpha }^2-280\alpha -400<0$

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

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

${25}^2-80-60+192<0$

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

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

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