Question

How do you write an equation of an ellipse in standard form given Length of major axis: 66, Vertices on the x-axis, Passes through the point (16.5 ,12 )?

Answer 1

`x^2/33^2+(3y^2)/24^2=1`

Explanation:

Let the half of length of major axis be a half of length of minor axis be b.

Given the length of major axis=66

So `2a=66=>a=33`

It is also given that the vertices are on x-axis.Let the coordinate of the center of ellipse be (c,0). Then the equation of ellipse may be written as.

`(x-c)^2/a^2+y^2/b^2=1`

The equation can be found out,if we consider c=0
So the equation becomes

`x^2/a^2+y^2/b^2=1....(1)`

Now a=33 and the equation passes through (16.5,12).So

`(16.5)^2/33^2+12^2/b^2=1`

`=>1/4+12^2/b^2=1`

`=>12^2/b^2=1-14=3/4`

`:.b^2=12^2*4/3=24^2/3`

Putting the values of `a^2 and b^2` in equation (1) we get

`x^2/33^2+(3y^2)/24^2=1`