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., xaxis) 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 coordinates 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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