Question

Which among the following points lies outside the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$.

• A. (4, 3)
• B. (4, 3.5)
• C. (9, 3)
• D. (7, 4)

Answer 1

The correct option is B

(4, 3.5)

For point (4,3)

$\frac{x^2}{16}-\frac{y^2}{9}=\frac{4^2}{16}-\frac{3^2}{9}$

$=0$

So (4,3) lies on the hyperbola

For point(9,3)

$\frac{x^2}{16}-\frac{y^2}{9}=\frac{9^2}{16}-\frac{3^2}{9}>0$

So the point(9,3) lies inside the hyperbola

Similarly the point (7,4) can also be shown to be situated inside the given hyperbola.

For point (4.3.5)

$\frac{x^2}{16}-\frac{y^2}{9}=\frac{4^2}{16}-\frac{{3.5}^2}{9}<1$.

A point $P(x_1,y_1)$ is outside the given parabola if $\frac{x^2}{16}-\frac{y^2}{9}<1$.

Here this is true only for point (4,3.5).

So the correct option is (b)