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
From Position of a Point with Respect to the Hyperbola to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Nov 29, 23 01:15 AM
Nov 29, 23 12:54 AM
Nov 29, 23 12:50 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.