A line parallel to the straight line 2x − y = 0 is tangent to the hyperbola x2/4 − y2/2 = 1 at the point (x1, y1).

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A line parallel to the straight line $2x-y=0$ is tangent to the hyperbola ${x^2\over4}-{y^2\over2}=1$ at the point $(x_1,\ y_1)$. Then $x_1^2+5y_1^2$ is equal to :

(1) 10

(2) 5

(3) 8

(4) 6
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Hyperbola topic is interesting

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verified

Ans. (4) 6

Sol. Tangent at $(x_1, y_1)$

$xx_1-2yy_1-4 = 0$

This is parallel to $2x-y = 0$

$\implies{x_1\over2y_1}=2$

$\implies x_1=4y_1\quad$.......(1)

Point $(x_1, y_1)$ lie on hyperbola.

${x_1^2\over4}-{y_1^2\over2}-1=0\quad$.......(2)

On solving eq. (1) and (2)

We get $x_1^2+5y_1^2=6$

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It was really hard question,
But you make it easy and interesting.