Area of Ellipse, what's the formula?

Geometry Level 4

There is an ellipse with 2 focal points

P = ( 0 , 2 ) P = (0, 2)

Q = ( 9 , 0 ) Q = (9, 0)

If one of the points on the perimeter of the ellipse is ( 4 , 0 ) (4, 0) , find the area of the ellipse.


The answer is 16.16.

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1 solution

Ronak Agarwal
Aug 21, 2014

I will be taking a a =Semi-major axis, b b =Semi-minor axis

We know area of ellipse = π a b =\pi ab . So we have to find a , b a,b

Now distance between two focal points = 2 a 2 b 2 =2\sqrt{{a}^{2}-{b}^{2}}

Applying distance formula we get :

9 2 + 2 2 = 85 = 2 a 2 b 2 \sqrt { { 9 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 85 }=2\sqrt {{a}^{2}-{b}^{2}}

So we get a 2 b 2 = 85 4 { a }^{ 2 }-{ b }^{ 2 }=\frac { 85 }{ 4 } (i)

Also we will use the property of ellipses that sum of distances from the two focal points on an ellipse is equal to the length of major axis = 2 a = 2a

Using this we get :

2 2 + 4 2 + 5 = 2 a \sqrt { { 2 }^{ 2 }+{ 4 }^{ 2 } } +5=2a

a = 5 + 2 5 2 \Rightarrow a=\frac { 5+2\sqrt { 5 } }{ 2 }

Using (i) we get :

b = 5 5 10 b=5\sqrt { 5 } -10

Put in the values of a , b a,b to get :

A r e a = 16.16 \boxed { Area=16.16 }

Same Way I did

Kunal Gupta - 6 years, 7 months ago

In fact, in a standard ellipse x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 and a > b a>b

S P = a + e x 1 + S P = a e x 1 S P + S P = 2 a \begin{aligned}SP &= a + ex_1\\+~~~~~~~~S'P &= a - ex_1\\\hline SP+S'P &= 2a\end{aligned} where S S and S S' are focii and P ( x 1 , y 1 ) P(x_1,y_1) is any point on the ellipse.

Kishore S. Shenoy - 5 years, 8 months ago

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Yep These Are The Expressions For Focal Distances Of Any General Point On The Ellipse Which Are Derived By Using Definition Of Ellipse(Locus Of Point Which moves such that ratio of its distance from a fixed point and a line remains constant which is Always Less than one where fixed point is the focii and fixed line is the corresponding directrix (x= a or -a for standard ellipse in which a>b)

Prakhar Bindal - 5 years, 7 months ago

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