**Area Of Ellipse**

If, you slice a cone with an inclined plane then the resulted oval shape is known as Ellipse. An ellipse is a generalized case of a closed conical section. There is a formula set to find **Area Of Ellipse**.

We have discussed various methods with examples of How to Find Area of Ellipse on this page of recruitmentresult.com. Students who want to take help of Ellipse Calculator to find Area Of Ellipse are advised to explore this page continuously.

Candidates, once you know the equation of the ellipse then it will become very easy to solve Area Of Ellipse. It is actually a very simple task but first you have to learn Formula Of Area Of Ellipse which has been well mentioned on our page.

**Area Of Ellipse**

__Definition of Ellipse__:

According to Kepler’s theory, “an ellipse is a curve in a plane surrounding two focal points such that the sum of the distances to the two focal points is constant for every point on the curve”.

**How to Find The Area Of Ellipse?**

The formula to find the area of ellipse is:

**A = π * a * b**

Where ‘a’ and ‘b’ are the lengths of the semi-major and semi-minor axes.

Calculate Area under Parabola: __Area of Parabola__

Let go through an **example** to understand briefly Area Of Ellipse Formula:

See the below given ellipse:

As you can see, the length semi-major axis of this ellipse is 8 inches, and the length of its semi-minor axis is 2 inches. Put these values into the Formula for Area of Ellipse. The value of π is constant that is 3.14:

A = π * a * b

A = 3.14 * 8 * 2

**A = 50.24**

What will be the result if we’re given an ellipse’s area and the length of one of its semi-axes?

In this case, we have area of the ellipse and the length of its semi-minor axis. Put these values into the formula:

A = π * a * b

157 = 3.14 * a * 5

157 = 15.7 * a

157 / 15.7 = a

10 = a

The length of the semi-major axis is **10 feet**.

Know How to Calculate TSA, CSA, LSA? __Surface Area of Cube Formula__

**Ellipse Calculator**

Ellipse calculator helps a lot to solve the equations of the ellipse. Students can solve it by putting the values of the problem and then hit on “Go” button. The resultant value will be the answer of Ellipse Calculator.

**For Example:**

**Problem 1:**** **

Given the following equation

9×2 + 4y2 = 36

- a) Find the x and y intercepts of the graph of the equation.
- b) Find the coordinates of the foci.
- c) Find the length of the major and minor axes.
- d) Sketch the graph of the equation.

**Solution to Example 1**

**a)**We first write the given equation in standard form by dividing both sides of the equation by 36

9×2 / 36 + 4y2 / 36 = 1

x2 / 4 + y2 / 9 = 1

x2 / 22 + y2 / 32 = 1

We now identify the equation obtained with one of the standard equation in the review above and we can say that the given equation is that of an ellipse with a = 3and b = 2 (NOTE: a >b) .

Set y = 0 in the equation obtained and find the x intercepts.

x2 / 22 = 1

Solve for x.

x2 = 22

x = ~+mn~ 2

Set x = 0 in the equation obtained and find the y intercepts.

Y^{2} / 32 = 1

Solve for y.

Y^{2} = 32

y = ~+mn~ 3

**b)**We need to find c first.

C^{2} = a^{2 }– b^{2}

‘a’ and ‘b’ were found in part a).

C^{2} = 3^{2} – 2^{2}

C^{2 }= 5

Solve for C.

c = ~+mn~ (5)^{1/2}

The foci are:

F1 (0 , (5)^{1/2}) and F2 (0 , -(5)^{1/2})

**C)** The major axis length is given by 2a = 6.

The minor axis length is given by 2b = 4.

**D)**Locate the x and y intercepts, find extra points if needed and sketch.

Get Derivation with Examples: __Area of Polygon Formula__

**Example 2:**

Find the equation of the normals at the end of the latus rectum of the ellipse x^{2}/a^{2} + y^{2}/b^{2}= c^{2} and find the condition when each normal through one end of the minor axis.

**Solution:**

The ellipse x^{2}/a^{2}+ y^{2}/b^{2} = c^{2}

⇒ x^{2}/(a^{2} c^{2} + y^{2}/(b^{2} c^{2}) = 1 …… (1)

Then the end point of the latus rectum is (ace,(b^{2} c)/a)

The normal at this point will be

(X-ace)/ (ace/ (a^{2} c^{2})) = (y-b^{2} c/a)/ ((b^{2} c/a)/ (b^{2} c^{2}))

⇒ ((x-ace))/e ac = (y-(b^{2} c)/a) ac

⇒ (x-ace)/e = y-(b^{2} c)/a

If this normal passes through (0, – bc), then, we have

(-ace)/e = -bc-b^{2}/a c

⇒ a = b + a(1 – e^{2})

⇒ b – ae^{2} = 0

⇒ b/1-e^{2} ⇒ b^{2}/a^{2} = e4

⇒ e4+ e^{2} = 1. This is required condition.

Get Formula of Circular Sector: __How to Find the Area of a Sector?__

__Final Note__:

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