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How to find the equation of an ellipse with foci and points?

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  • Last Updated : 01 Feb, 2022

When a plane crosses a cone, conic sections, also known as conics, are created. The geometry of these sections is determined by the angle at which they cross. As a result, conic sections are divided into four categories: circle, ellipse, parabola, and hyperbola. Each of these forms has its own set of mathematical features and equations. The ellipse is discussed below.

Ellipse

As a conic section, an ellipse is a shape formed when a plane intersects a cone at an angle (β) less than the right angle but more than the angle formed at the vertex of a cone (α). In other words, an ellipse is formed when a plane cuts a cone at an angle β such that α<β<90o.

As shown in the figure above, a cone and a plane intersect at an angle β which is less than the right angle but more than α to form an ellipse due to the intersection.

Equation of an ellipse

  • The standard equation of an ellipse centred at (h, k) with a major axis parallel to the x-axis is given by:

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

where the coordinates of the vertex are (h±a, 0), coordinates of co-vertex are (h, k±b) and the coordinates of foci are (h±c, k), where c2 = a2 – b2.

Horizontal ellipse

  • The standard equation of an ellipse centred at (h, k) with a major axis parallel to the y-axis is given by:

\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1 , 

where the coordinates of the vertex are (h, k±a), coordinates of co-vertex are (h±b, k) and the coordinates of foci are (h, k±c), where  c2 = a2 – b2.

Vertical ellipse

How to find the equation of an ellipse with foci and points?

Solution:

To find the equation of an ellipse, we need the values a and b. Now, it is known that the sum of the distances of a point lying on an ellipse from its foci is equal to the length of its major axis, 2a. The value of a can be calculated by this property. To calculate b, use the formula c2 = a2 – b2. Substitute the obtained values of a and b in the standard form to get the required equation.

Let us understand this method in more detail through an example. 

Example: Say, an ellipse passing through the origin with foci (±4, 0) and point (–4, 1.8).

Using the formula, we have

2a=1.8+\sqrt{(4-(-4))^2+(0-1.8)^2}

2a = 10

a = 5

Put a = 5 in c2 = a2 – b2 to find b.

b2 = 25 – 16

b2 = 9 

As the ellipse lies on x-axis, the equation is of the form \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1   .

So, the equation is, \frac{x^2}{25}+\frac{y^2}{9}=1   .

Similar Problems

Question 1. Find the equation of an ellipse passing through the origin with foci (±7, 0) and point (6, 2).

Solution:

Using the formula, we have

2a=\sqrt{(7-6)^2+(0-2)^2}+\sqrt{(-7-6)^2+(0-2)^2}\\ 2a=\sqrt{5}+\sqrt{173}\\

2a = 15.74

a = 7.87

Put a = 7.87 in c2 = a2 – b2 to find b.

b2 = 62 – 49

b2 = 13

As the ellipse lies on x-axis, the equation is of the form \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1   .

So, the equation is, \frac{x^2}{62}+\frac{y^2}{13}=1   .

Question 2. Find the equation of an ellipse passing through the origin with foci (±5√3, 0) and point (6, 4).

Solution:

Using the formula, we have

2a=\sqrt{(5\sqrt{3}-6)^2+(0-4)^2}+\sqrt{(-5\sqrt{3}-6)^2+(0-4)^2}

2a = 20

a = 10

Put a = 10 in c2 = a2 – b2 to find b.

b2 = 100 – 75

b2 = 25

As the ellipse lies on x-axis, the equation is of the form \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1   .

So, the equation is, \frac{x^2}{100}+\frac{y^2}{25}=1   .

Question 3. Find the coordinates of the major axis of the ellipse with foci (0, ±5) and minor axis (12, 0).

Solution:

We have, c = 5 and b = 12.

Put these in c2 = a2 – b2 to find a.

a2 = 122 + 52

a2 = 169

a = 13

The coordinates of major axis are (0, ±13).

Question 4. Find the equation of the ellipse passing through the origin if a = 3, b = 5, and the major axis is parallel to the x-axis.

Solution:

The major axis is parallel to x-axis, so the ellipse lies on x-axis.

The equation is of the form \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1   .

Here, a = 3, b = 5, h = 0 and k = 0.

So, the equation becomes,

\frac{(x-0)^2}{3^2}+\frac{(y-0)^2}{5^2}=1

\frac{x^2}{9}+\frac{y^2}{25}=1

Question 5. Find the equation of the ellipse passing through the origin if a = 13, b = 5, and the minor axis is parallel to the x-axis.

Solution:

The major axis is parallel to x-axis, so the ellipse lies on x-axis.

The equation is of the form \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1

Here a = 13 and b = 5, h = 0 and k = 0.

So, the equation becomes,

\frac{(x-0)^2}{5^2}+\frac{(y-0)^2}{13^2}=1

\frac{x^2}{25}+\frac{y^2}{169}=1


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