Question

Which of the following point lies outside the hyperbola $7x^2-2y^2+12xy-2x+14y-22=0$

• A. $(1,2)$
• B. $(2,3)$
• C. $(3,3)$
• D. $(1,0)$

Answer 1

The correct option is D $(1,0)$

Equation of given Hyperbola is $7x^2-2y^2+12xy-2x+14y-22=0$

Let $S=7x^2-2y^2+12xy-2x+14y-22$

At any given point $P(x_1,y_1)$ $\to S_1=7x_1^2-2y_1^2+12x_1{y}_1-2x_1+14y_1-22$

Now according to Properties of hyperbola if $S_1<0$ then point $P(x_1,y_1)$ lies outside the hyperbola.

if $S_1=0$ then point $P(x_1,y_1)$ lies on the hyperbola.

and if $S_1>0$ then point $P(x_1,y_1)$ lies inside the hyperbola.

Now for point $(1,2)$, $S_{(1,2)}=27......(>0)$

Now for point $(2,3)$, $S_{(2,3)}=98......(>0)$

Now for point $(3,3)$, $S_{(3,3)}=167......(>0)$

Now for point $(1,0)$, $S_{(1,0)}=-17......(<0)$

So we can see from all these points only $(1,0)$ point gives a negative value of $S_1$,

$\because$ $S_{(1,0)}<0$, Hence point $(1,0)$ lies outside of the given hyperbola.

Because rest points give a positive value of $S_1$, that's why they lie inside of the given hyperbola.

So correct option is $D$.