# Position of a Point with Respect to the Hyperbola

We will learn how to find the position of a point with respect to the hyperbola.

The point P (x$$_{1}$$, y$$_{1}$$) lies outside, on or inside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 according as $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ – 1 < 0, = or > 0.

Let P (x$$_{1}$$, y$$_{1}$$) be any point on the plane of the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 ………………….. (i)

From the point P (x$$_{1}$$, y$$_{1}$$) draw PM perpendicular to XX' (i.e., x-axis) and meet the hyperbola at Q.

According to the above graph we see that the point Q and P have the same abscissa. Therefore, the co-ordinates of Q are (x$$_{1}$$, y$$_{2}$$).

Since the point Q (x$$_{1}$$, y$$_{2}$$) lies on the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1.

Therefore,

$$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{2}^{2}}{b^{2}}$$ = 1

$$\frac{y_{2}^{2}}{b^{2}}$$ = $$\frac{x_{1}^{2}}{a^{2}}$$ - 1 ………………….. (i)

Now, point P lies outside, on or inside the hyperbola according as

PM <, = or > QM

i.e., according as y$$_{1}$$ <, = or > y$$_{2}$$

i.e., according as $$\frac{y_{1}^{2}}{b^{2}}$$ <, = or > $$\frac{y_{2}^{2}}{b^{2}}$$

i.e., according as $$\frac{y_{1}^{2}}{b^{2}}$$ <, = or > $$\frac{x_{1}^{2}}{a^{2}}$$ - 1, [Using (i)]

i.e., according as $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ <, = or > 1

i.e., according as $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 <, = or > 0

Therefore, the point

(i) P (x$$_{1}$$, y$$_{1}$$) lies outside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 if PM < QM

i.e., $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 < 0.

(ii) P (x$$_{1}$$, y$$_{1}$$) lies on the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 if PM = QM

i.e., $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 = 0.

(ii) P (x$$_{1}$$, y$$_{1}$$) lies inside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 if PM < QM

i.e., $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 > 0.

Hence, the point P(x$$_{1}$$, y$$_{1}$$) lies outside, on or inside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 according as x$$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$  - 1 <, = or > 0.

Note:

Suppose E$$_{1}$$ = $$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1, then the point P(x$$_{1}$$, y$$_{1}$$) lies outside, on or inside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 according as E$$_{1}$$ <, = or > 0.

Solved examples to find the position of the point (x$$_{1}$$, y$$_{1}$$) with respect to an hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1:

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

Solution:

We know that the point (x$$_{1}$$, y$$_{1}$$) lies outside, on or inside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 according as

$$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ – 1 < , = or > 0.

For the given problem we have,

$$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 = $$\frac{2^{2}}{9}$$ - $$\frac{(-3)^{2}}{25}$$ – 1 = $$\frac{4}{9}$$ - $$\frac{9}{25}$$ - 1 = - $$\frac{206}{225}$$ < 0.

Therefore, the point (2, - 3) lies outside the hyperbola $$\frac{x^{2}}{9}$$ - $$\frac{y^{2}}{25}$$ = 1.

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

Solution:

We know that the point (x$$_{1}$$, y$$_{1}$$) lies outside, on or inside the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 according as

$$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 <, = or > 0.

For the given problem we have,

$$\frac{x_{1}^{2}}{a^{2}}$$ - $$\frac{y_{1}^{2}}{b^{2}}$$ - 1 = $$\frac{3^{2}}{9}$$ - $$\frac{(-4)^{2}}{16}$$ - 1 = $$\frac{9}{9}$$ - $$\frac{16}{16}$$ - 1 = 1 - 1 - 1 = -1 < 0.

Therefore, the point (3, - 4) lies outside the hyperbola $$\frac{x^{2}}{9}$$ - $$\frac{y^{2}}{16}$$ = 1.

The Hyperbola