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$(x_1,y_1)$.

Then ${x_1}^2+5{y_1}^2$ is equal to

• A. $6$
• B. $10$
• C. $8$
• D. $5$

Answer 1

The correct option is A

$6$

The explanation for the correct option:

Step 1: Expressing the Given information:

$\begin{array}{rcl}T:\left(\frac{x{x}_1}{4}\right)-\left(\frac{y{y}_1}{2}\right)& =& 1..\left(1\right)\\ t:2x-y& =& 0\\ T:2x-y& =& \lambda ..\left(2\right)\end{array}$

Step 2: Finding $\lambda$

Now, compare (1) and (2)

$\begin{array}{rcl}{x}_1÷\frac{4}{2}& =& y_1÷\frac{2}{1}\\ & =& \frac{1}{\lambda }\end{array}$

$\begin{array}{rcl}{x}_1& =& \frac{8}{\lambda },y_1=\frac{2}{\lambda }\end{array}$

(x₁, y₁) lies on hyperbola

$\begin{array}{rcl}\left(\frac{64}{4{\lambda }^2}\right)-\left(\frac{4}{2{\lambda }^2}\right)& =& 1\\ 14& =& {\lambda }^2\end{array}$

Step 3:Finding the value of ${x_1}^2+5{y_1}^2$

Now

$\begin{array}{rcl}{x_1}^2+5{y_1}^2& =& \left(\frac{64}{{\lambda }^2}\right)+5\left(\frac{4}{{\lambda }^2}\right)\\ & =& \left(\frac{64}{14}\right)+\left(\frac{20}{14}\right)\\ & =& \frac{84}{14}\\ & =& 6\end{array}$

$\therefore$${x_1}^2+5{y_1}^2$ is equal to $\mathbf{6}$

Hence, Option ‘A’ is Correct.