Question

A line parallel to the straight line $2x-y=0$ is tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$ at the point $\left(x_1,y_1\right)$. Then $x_1^2+5y_1^2$ is equal to

Answer 1

Answer: 6

Equation by straight line $2x-y=0$;
Equation by tangent $\frac{x{x}_1}{4}-\frac{y{y}_1}{2}=1$
Both are parallel
$\frac{x_1}{y_1}=4$............(1) equation
Now at the point $\left(x_1{y}_1\right)$ lied on given hyperbola
$\therefore \frac{x_1^2}{4}-\frac{y_1^2}{2}-1=0$...........(2) equation
On solving (i) and (ii) we get $y_1=\pm \sqrt{\frac{2}{7}}$ and $x_1=\pm \sqrt{\frac{32}{7}}$
$x_1^2+5y_1^2=\frac{32}{7}+5\times \frac{2}{7}=\frac{32+10}{7}=\frac{42}{7}=6$