Question

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

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

Answer 1

The correct option is D

$(4,-\sqrt{6})$

There is a set of questions in which we can compare 2 equations representing the same curve and find the parameters. This question is one among them.

We are asked the point of contact. We know the standard form of a tangent at a point $(x_1,y_1)$. This equation can be compared with the given equation of tangent to

find out the point of contact.

The given hyperbola is,

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

the tangent at a point $(x_1,y_1)$ is,

$x{x}_1-2y{y}_1-4=0$

Now we compare it with the given tangent which is said to be the same after making the constant term the same by multiplying with a constant number if needed.

$2x+\sqrt{6}y-2=0$

i.e.,$4x+2\sqrt{6}y-4=0$

$x_1=4$

$-2y_1=2\sqrt{6}$

$y_1=-\sqrt{6}$

$\therefore (x_1,y_1)=(4,-\sqrt{6})$