# A line parallel to the straight line 2 x-y=0 is tangent to the hyperbola

Question:

A line parallel to the straight line $2 x-y=0$ is

tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point

$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$. Then $\mathrm{x}_{1}^{2}+5 \mathrm{y}_{1}^{2}$ is equal to :

1. 5

2. 6

3. 8

4. 10

Correct Option: , 2

Solution:

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

$\frac{x x_{1}}{4}-\frac{y y_{1}}{2}=1 \quad(T=0)$

Slope : $\frac{1}{2} \frac{x_{1}}{y_{1}}=2 \Rightarrow x_{1}=4 y_{1}$...(1)

$\left(x_{1}, y_{1}\right)$ lies on hyperbola

$\Rightarrow \frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1$ ........(2)

From (1) & (2)

$\frac{\left(4 y_{1}\right)^{2}}{4}-\frac{y_{1}^{2}}{2}=1 \Rightarrow 4 y_{1}^{2}-\frac{y_{1}^{2}}{2}=1$

$\Rightarrow 7 \mathrm{y}_{1}^{2}=2 \Rightarrow \mathrm{y}_{1}^{2}=2 / 7$

Now $x_{1}^{2}+5 y_{1}^{2}=\left(4 y_{1}\right)^{2}+5 y_{1}^{2}$

$=(21) y_{1}^{2}=21 \times \frac{2}{7}=6$