Question

Determine the position of $(4,10)$ to the hyperbola $\frac{x^2}{16}-\frac{y^2}{20}=1$.

• A. It lies outside the hyperbola
• B. It lies inside the hyperbola
• C. It lies on the periphery of the hyperbola
• D. None of these

Answer 1

The correct option is A It lies outside the hyperbola

The given Hyperbola is $\frac{x^2}{16}-\frac{y^2}{20}=1$ or $\frac{x^2}{16}-\frac{y^2}{20}-1=0$

Let $S=\frac{x^2}{16}-\frac{y^2}{20}-1$

Then any point $P(x_1,y_1)$ will lie inside, on or the outside of the hyperbola $\frac{x^2}{16}-\frac{y^2}{20}=1$ depending upon the value of $S_P=(\frac{x_1^2}{16}-\frac{y_1^2}{20}-1)$

If the value of $S_P>0$ then point $P(x_1,y_1)$ lies inside of the hyperbola.

If the value of $S_P=0$ then point $P(x_1,y_1)$ lies on the perphery of the hyperbola.

If the value of $S_P<0$ then point $P(x_1,y_1)$ lies outside of the hyperbola.

Now let the value of the term $S_P=\frac{x_1^2}{16}-\frac{y_1^2}{20}-1$ at given point $P(4,10)$ is $S_{(4,10)}$

Then $S_{(4,10)}$ $=\frac{4^2}{16}-\frac{(10{)}^2}{20}-1$

$\Rightarrow$ $1-5-1=-5.....(<0)$

As value of term $S_P=\frac{x_1^2}{16}-\frac{y_1^2}{20}-1$ at point $P(4,10)$ is less than $0$, Hence the point lies outside of the hyperbola.

So correct option is $A$