# Thread: ellipse 1

1. ## ellipse 1

An ellipse have the centre in origin and axis lower on the OX axis. Find your equation knowing that the major axis is twice of the minor axis and the ellipse passing through point P $(\frac{\sqrt{7}}{2},3)$

$16x^2+4y^2=64$

2. Originally Posted by Apprentice123
An ellipse have the centre in origin and axis lower on the OX axis. Find your equation knowing that the major axis is twice of the minor axis and the ellipse passing through point P $(\frac{\sqrt{7}}{2},3)$

$16x^2+4y^2=64$
It took me a little bit more playing before I understood your "axis lower on the OX axis". The book answer did help.

You probably meant that the major axis is along the y-axis.

So, with a = semi-major axis and b = semi-minor axis,
(x^2)/(b^2) +(y^2)/(a^2) = 1 -----(i)

a = 2b ----given
a^2 = 4b^2

At point (sqrt(7) /2,3),
Substitutions into (i),
[(Sqrt(7) /2)^2]/(b^2) + (3^2)/(4b^2) = 1
7/(4b^2) +9/(4b^2) = 1
Clear the fractions, multiply both sides by 4b^2,
7 + 9 = 4b^2
b = 2 units --------**
And so, a = 2b = 4 units ----**

Hence, the ellipse is
(x^2)/(2^2) +(y^2)/(4^2) = 1
(x^2)/4 +(y^2)/16 = 1
Clear the fractions, multiply both sides by 16,
4x^2 +y^2 = 16 -----------------------------answer.

The book answer is 16x^2 +4y^2 = 64.
The book answer is on steroids.

3. thank you