Question

If $P(x_1,y_1)$ is such that $\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}>1$.

Then the point P situates outside the standard hyperbola $\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}>=1$.

• A. True
• B. False

Answer 1

The correct option is A

True

The condition for a point to situate with respect to a hyperbola is as below

$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}<1$ implies p outside

$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$ implies p on the hyperbola

$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}>1$ implies p inside the hyperbola

For remembering its better to see what happens if we substitude (0,0)

in the curve. Note that (0,0) is outside the curve.

If any other point yields the same result then that point situates on

the same side as that of the orign

Proof

If p is the point under considerration. Drop perpendicular from P to the x-axis.

this lines meets the hyperbola at $Q(x_1,y_2)$.

$PR>QR$

Is $y_1>y_2$

$\frac{y_1^2}{b^2}>\frac{y_2^2}{b^2}$

$-\frac{y_1^2}{b^2}<-\frac{y_2^2}{b^2}$

$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}<\frac{x_1^2}{a^2}-\frac{y_2^2}{b^2}$

i.e., $\frac{x_1}{a^2}-\frac{y_1^2}{b^2}<1$ since $(x_1,y_1)$ lies ON the hyperbola