Expert: Ahmed Salami - 10/5/2004

Question
Find the standard form equation of the conic.

An ellipse with minor axis along the x axis, one vertex at ( -1,10) and etrenticity =4/5

Answer
Hi Anna,
Sorry about the delay.
The standard form of the equation of an ellipse is written
x^2/a^2 + y^2 /b^2 = 1
where a and b are the lengths of the semi axis that coincides with the x and y axes respectively. The equation is used when the centre of the ellipse is at the origin(0,0). If the origin is at(h,k), the equation becomes
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
Now, eccentricity = sqrt[1 - (b/a)^2]or sqrt[1 - (a/b)^2]
whichever gives the fraction less than 1.
In this situation, the minor axis is along the x axis i.e b > a and e = 4/5
Therefore,
sqrt[1 - (a/b)^2]= 4/5
[1 - (a/b)^2]= 16/25
(a/b)^2 = 9/25
a/b = 3/5
but b = 10, then a = 6
So the equation is
(x+1)^2/6^2 + y^2/10^2 = 1
multiplying through by 3600
100(x+1)^2 + 36y^2 = 3600
dividing by 4 gives
25(x+1)^2 + 9y^2 = 900
9y^2 + 25(x^2 + 2x + 1)= 900
9y^2 + 25x^2 + 50x + 25 = 900
9y^2 + 25x^2 + 50x - 875 = 0

I hope you get it.
Regards.