how to find the equation of an ellipse

Ellipse is an important topic in the conic section. Information technology is the ready of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. These stock-still points are chosen foci of the ellipse. In this article, we will learn how to find the equation of ellipse when given foci.

The centre of the ellipse is the midpoint of the line segment joining the foci of the ellipse. The major axis is the line segment passing through the foci of the ellipse. The line segment passing through the middle and perpendicular to the major axis is chosen the minor axis. The endpoints of the major centrality are called the vertices of the ellipse. The distance between the foci is denoted by 2c. And then the length of the semi-major axis is a and the semi-pocket-sized axis is b. The length of the major centrality is denoted past 2a and the pocket-sized centrality is denoted past 2b.

The relation between the semi-major axis, semi-small-scale axis and the distance of the focus from the centre of the ellipse is given by the equation c = √(a² – b²).

The standard equation of ellipse is given past (x²/a²) + (y^two/b²) = i.

The foci ever lie on the major axis. The major axis tin exist known by finding the intercepts on the axes of symmetry, i.e, the major centrality is along the x-axis if the coefficient of 10² has the larger denominator and it is along the y-axis if the coefficient of y² has the larger denominator.

Steps to notice the Equation of the Ellipse.

1. Find whether the major centrality is on the x-centrality or y-axis.

2. If the coordinates of the vertices are (±a, 0) and foci is (±c, 0), then the major axis is parallel to ten axis. So use the equation

\(\begin{array}{l}\frac{x^{2}}{a^{2}}+\frac{y^{two}}{b^{2}} = 1\end{array} \)

.

3. If the coordinates of the vertices are (0, ±a) and foci is (0,±c), and so the major axis is parallel to y axis. Then use the equation

\(\brainstorm{assortment}{l}\frac{x^{ii}}{b^{2}}+\frac{y^{2}}{a^{two}} = 1\finish{array} \)

.

four. Using the equation c^two = (aⁱⁱ – b²), find b².

5. Substitute the values of a² and b² in the standard form.

Solved Examples

Example one.

Notice the equation of an ellipse with vertices (0, ±8) and foci (0, ±iv).

Solution:

Given vertices (0, ±8) and foci (0, ±4).

Here a = 8 and c = 4

c² = a² – b²

Then b² = a^two – c²

= eight² – 4²

= 64 – 16

= 48

We use the equation

\(\begin{array}{l}\frac{10^{ii}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\finish{assortment} \)

.

So (x²/48) + y²/64 = ane is the required equation.

Example two.

Find the equation of the ellipse that has vertices (± 13, 0) and foci are (± v, 0).

Solution:

Given vertices (± thirteen, 0) and foci are (± v, 0).

Vertices are in 10-axis. And so the equation volition be of the course (ten^two/a²) + (y²/b²) = 1.

Here a = thirteen and c = 5

bⁱⁱ = a² – c²

b² = 13² – 5²

b² = 144

So (x²/169) + y²/144 = one is the required equation.

Example 3: Write an equation for the ellipse having foci at (–2, 0) and (2, 0) and eccentricity east = iii/4.

Solution:

Given foci is (-2,0) and (ii,0).

And so major centrality is parallel to x axis. Hence we use the equation (x ² /a ⁱⁱ )+(y ² /b ² ) = 1.

Given e = 3/iv and c= 2

we know, e = c/a

So a = c/e = eight/3

a ² = 64/ix

b ² = a ² -c ⁱⁱ

=(64/ix)-4

= (64-36)/9

= 28/9

The equation of ellipse is (ten ² /a ^two )+(y ² /b ⁱⁱ ) = 1

(9x ² /64)+(9y ² /28)=1 is the required equation.

Related Links:

• Conic Sections
• How to Find Equation of Ellipse with Foci and Major Centrality
• Conic Sections Previous Year Questions With Solutions

Ellipse and Hyperbola – JEE Important and Previous Twelvemonth Questions

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