Question

If the line $2x+\sqrt{6}y=2$ touches the hyperbola $x^2-2y^2=4,$ then the point of contact is

• A. $(-2,\sqrt{6})$
• B. $(-5,2\sqrt{6})$
• C. $(\frac{1}{2},\frac{1}{\sqrt{6}})$
• D. $(4,-\sqrt{6})$

Answer 1

The correct option is D $(4,-\sqrt{6})$

Line is $2x+\sqrt{6}y=2$

Hyperbola is $x^2-2y^2=4$

$\Rightarrow x^2=2y^2+4$

$\Rightarrow 4x^2=8y^2+16$

$\Rightarrow (2x{)}^2=8y^2+16$ ___(1)

from line $2x=(2-\sqrt{6}y)$ ___ (2)

putting the value of 2x in equation (1)

$\Rightarrow (2-\sqrt{6}y{)}^2=8y^2+16$

$\Rightarrow 4+6y^2-4\sqrt{6}y=8y^2+16$

$\Rightarrow 0=2y^2+4\sqrt{6}y+12$

divide by 2

$\Rightarrow y^2+2\sqrt{6}y+6=0$

$\Rightarrow (y+\sqrt{6}{)}^2=0$

$\Rightarrow y=-\sqrt{6}$

From (2)

$2x=2-\sqrt{6}(-\sqrt{6})$

$=2+6=+8$

$\Rightarrow x=4$

$\Rightarrow$ point is $(4,-\sqrt{6})$