Question

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

• A. inside
• B. outside
• C. on the periphery
• D. Cannot be determined

Answer 1

The correct option is C on the periphery

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

Let $S=\frac{x^2}{9}-\frac{y^2}{16}-1$

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 $S_P=(\frac{x_1^2}{9}-\frac{y_1^2}{16}-1)$

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 $S_P=\frac{x_1^2}{9}-\frac{y_1^2}{16}-1$ at given point $P(3sec\theta ,4tan\theta )$ is $S_{(3sec\theta ,4tan\theta )}$

Then $S_{(3sec\theta ,4tan\theta )}$ $=\frac{(3sec\theta {)}^2}{9}-\frac{(4tan\theta {)}^2}{16}-1$

$\Rightarrow$ $se{c}^2\theta -ta{n}^2\theta -1=1-1=0.....=\left(0\right)$

As value of term $S_P=\frac{x_1^2}{9}-\frac{y_1^2}{16}-1$ at point $P(3sec\theta ,4tan\theta )$ is equal to $0$, Hence the point lies on the periphery of the hyperbola.

So correct option is $C$