# 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 )?

Jul 28, 2016

${x}^{2} / {33}^{2} + \frac{3 {y}^{2}}{24} ^ 2 = 1$

#### 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 $2 a = 66 \implies a = 33$

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.

${\left(x - c\right)}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

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

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1. \ldots \left(1\right)$

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

${\left(16.5\right)}^{2} / {33}^{2} + {12}^{2} / {b}^{2} = 1$

$\implies \frac{1}{4} + {12}^{2} / {b}^{2} = 1$

$\implies {12}^{2} / {b}^{2} = 1 - 14 = \frac{3}{4}$

$\therefore {b}^{2} = {12}^{2} \cdot \frac{4}{3} = {24}^{2} / 3$

Putting the values of ${a}^{2} \mathmr{and} {b}^{2}$ in equation (1) we get

${x}^{2} / {33}^{2} + \frac{3 {y}^{2}}{24} ^ 2 = 1$