Position of a Point with respect to the Ellipse

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.



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.

Share this page: What’s this?

Recent Articles

  1. Word Problems on Area and Perimeter | Free Worksheet with Answers

    Jul 26, 24 04:58 PM

    word problems on area and perimeter

    Read More

  2. Worksheet on Perimeter | Perimeter of Squares and Rectangle | Answers

    Jul 26, 24 04:37 PM

    Most and Least Perimeter
    Practice the questions given in the worksheet on perimeter. The questions are based on finding the perimeter of the triangle, perimeter of the square, perimeter of rectangle and word problems. I. Find…

    Read More

  3. Perimeter and Area of Irregular Figures | Solved Example Problems

    Jul 26, 24 02:20 PM

    Perimeter of Irregular Figures
    Here we will get the ideas how to solve the problems on finding the perimeter and area of irregular figures. The figure PQRSTU is a hexagon. PS is a diagonal and QY, RO, TX and UZ are the respective d…

    Read More

  4. Perimeter and Area of Plane Figures | Definition of Perimeter and Area

    Jul 26, 24 11:50 AM

    Perimeter of a Triangle
    A plane figure is made of line segments or arcs of curves in a plane. It is a closed figure if the figure begins and ends at the same point. We are familiar with plane figures like squares, rectangles…

    Read More

  5. 5th Grade Math Problems | Table of Contents | Worksheets |Free Answers

    Jul 26, 24 01:35 AM

    In 5th grade math problems you will get all types of examples on different topics along with the solutions. Keeping in mind the mental level of child in Grade 5, every efforts has been made to introdu…

    Read More