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 :

• A. 5
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (B)

Slope of tangent is $2,$ Tangent of hyperbola
$\frac{x^2}{4}-\frac{y^2}{2}=1$ at the point $\left(x_1,y_1\right)$ is
$\frac{x{x}_1}{4}-\frac{y{y}_1}{2}=1(T=0)$
Slope : $\frac{1}{2}\frac{x_1}{y_1}=2$
$\Rightarrow x_1=4y_1\dots$(1)
$\left(x_1,y_1\right)$ lies on hyperbola
$\Rightarrow \frac{x_1^2}{4}-\frac{y_1^2}{2}=1\dots$(2)
From (1)$\&(2)$
$\frac{{\left(4y_1\right)}^2}{4}-\frac{y_1^2}{2}=1$
$\Rightarrow 4y_1^2-\frac{y_1^2}{2}=1$
$\Rightarrow 7y_1^2=2$
$\Rightarrow y_1^2=2/7$
Now $x_1^2+5y_1^2={\left(4y_1\right)}^2+5y_1^2$
$=(21)y_1^2=21\times \frac{2}{7}=6$