Question

Determine the position of the point $(2,-3)$ with respect to the hyperbola $\frac{x^2}{9}-\frac{y^2}{25}=1$

• A. Inside the hyperbola
• B. Outside the hyperbola
• C. On the hyperbola
• D. Cannot be determined from the given information.

Answer 1

The correct option is B Outside the hyperbola

The given Hyperbola is $\frac{x^2}{9}-\frac{y^2}{25}=1$ or $\frac{x^2}{9}-\frac{y^2}{25}-1=0$

Then any point $P(x_1,y_1)$ will lie inside, on or the outside of the hyperbola $\frac{x^2}{9}-\frac{y^2}{25}=1$ depending upon the value of $(\frac{x_1^2}{9}-\frac{y_1^2}{25}-1)$

If the value of $\frac{x_1^2}{9}-\frac{y_1^2}{25}-1>0$ then point $P(x_1,y_1)$ lies inside of the hyperbola.

If the value of $\frac{x_1^2}{9}-\frac{y_1^2}{25}-1=0$ then point $P(x_1,y_1)$ lies on the periphery of the hyperbola.

If the value of $\frac{x_1^2}{9}-\frac{y_1^2}{25}-1<0$ then point $P(x_1,y_1)$ lies outside of the hyperbola.

Now let the value of the term $\frac{x_1^2}{9}-\frac{y_1^2}{25}-1$ at given point $(2,-3)$ is $(\frac{x_1^2}{9}-\frac{y_1^2}{25}-1{)}_{(2,-3)}$

Then $(\frac{x_1^2}{9}-\frac{y_1^2}{25}-1{)}_{(2,-3)}=\frac{2^2}{9}-\frac{(-3{)}^2}{25}-1$

$\to$ $\frac{100-81-225}{225}=\frac{-206}{225}<0$

As value of term $\frac{x_1^2}{9}-\frac{y_1^2}{25}-1$ at point $P(2,-3)$ is less than $0$, Hence the point lies outside of the hyperbola.

So correct option is $B$