We will learn how to find the position of a point with respect to the ellipse.
The point P (x\(_{1}\), y\(_{1}\)) lies outside, on or inside the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 ………………….. (i)
From the point P (x\(_{1}\), y\(_{1}\)) draw PM perpendicular to XX' (i.e., x-axis) and meet the ellipse 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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1.
Therefore,
\(\frac{x_{1}^{2}}{a^{2}}\) + \(\frac{y_{2}^{2}}{b^{2}}\) = 1
\(\frac{y_{2}^{2}}{b^{2}}\) = 1 - \(\frac{x_{1}^{2}}{a^{2}}\) ………………….. (i)
Now, point P lies outside, on or inside the ellipse 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 < 1 - \(\frac{x_{1}^{2}}{a^{2}}\), [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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 according as E\(_{1}\) >, = or < 0.
Solved examples to find the position of the point (x\(_{1}\), y\(_{1}\)) with respect to an ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1:
1. Determine the position of the point (2, - 3) with respect to the ellipse \(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{25}\) = 1.
Solution:
We know that the point (x\(_{1}\), y\(_{1}\)) lies outside, on or inside the ellipse
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 = - \(\frac{44}{225}\) < 0.
Therefore, the point (2, - 3) lies inside the ellipse \(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{25}\) = 1.
2. Determine the position of the point (3, - 4) with respect to the ellipse \(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{16}\) = 1.
Solution:
We know that the point (x\(_{1}\), y\(_{1}\)) lies outside, on or inside the ellipse
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 ellipse \(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{16}\) = 1.
● The Ellipse
11 and 12 Grade Math
From Position of a Point with respect to the Ellipse 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.
Dec 11, 24 09:08 AM
Dec 09, 24 10:39 PM
Dec 09, 24 01:08 AM
Dec 08, 24 11:19 PM
Dec 07, 24 03:38 PM
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.