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






11 and 12 Grade Math 

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