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  • A line parallel to the straight line is tangent to the hyperbola <img alt="\mathrm{\f

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

Option: 1

6


Option: 2

8


Option: 3

10


Option: 4

3


Answers (1)

best_answer

The given equation of hyperbola is \mathrm{\frac{x^2}{4}-\frac{y^2}{2}=1} and equation of straight line is \mathrm{2 x-y=0.}

Equation of tangent at \mathrm{\left(x_1, y_1\right)} is \mathrm{\frac{x \cdot x_1}{4}-\frac{y \cdot y_1}{2}=1}

\mathrm{\Rightarrow x \cdot x_1-2 \cdot y \cdot y_1-4=0} and this is parallel to \mathrm{2 x-y=0.}

\mathrm{\therefore \frac{x_1}{2 y_1}=2 \Rightarrow x_1-4 y_1=0 }
Now as the point \mathrm{\left(x_1, y_1\right)} lies on given hyperbola.

\mathrm{\therefore \frac{x_1^2}{4}-\frac{y_1^2}{2}-1=0}

On solving (i) and (ii), we get \mathrm{y_1= \pm \sqrt{\frac{2}{7}}} and

\begin{aligned} &\mathrm{ x_1= \pm \sqrt{\frac{32}{7}} }\\ &\mathrm{ \therefore x_1^2+5 y_1^2=\frac{32}{7}+5 \times \frac{2}{7}=\frac{32+10}{7}=\frac{42}{7}=6}\\ \end{aligned}


 

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