Question

Determine the position of the point (3, - 4) with respect to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$

• A. Inside
• B. Outside
• C. on the periphery
• D. None of these

Answer 1

The correct option is B Outside

The given Hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$ or $\frac{x^2}{9}-\frac{y^2}{16}-1=0$

Then any point $P(x_1,y_1)$ will lie inside, on or the outside of the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ depending upon the value of $(\frac{x_1^2}{9}-\frac{y_1^2}{16}-1)$

If the value of $\frac{x_1^2}{9}-\frac{y_1^2}{16}-1>0$ then point $P(x_1,y_1)$ lies inside of the hyperbola.

If the value of $\frac{x_1^2}{9}-\frac{y_1^2}{16}-1=0$ then point $P(x_1,y_1)$ lies on the hyperbola.

If the value of $\frac{x_1^2}{9}-\frac{y_1^2}{16}-1<0$ then point $P(x_1,y_1)$ lies outside of the hyperbola.

Now let the value of the term $\frac{x_1^2}{9}-\frac{y_1^2}{16}-1$ at given point $(3,-4)$ is ${(\frac{x_1^2}{9}-\frac{y_1^2}{16}-1)}_{(3,-4)}$

Then ${(\frac{x_1^2}{9}-\frac{y_1^2}{16}-1)}_{(3,-4)}=\frac{3^2}{9}-\frac{(-4{)}^2}{16}-1$

$\to$ $1-1-1=-1.....(<0)$

As value of term $\frac{x_1^2}{9}-\frac{y_1^2}{16}-1$ at point $P(3,-4)$ is less than $0$, Hence the point lies outside of the hyperbola.

So correct option is $B$