# Question #d2701

Aug 20, 2016

$\textcolor{red}{{\left(x - 4\right)}^{2} / {3}^{2} + {\left(y + 1\right)}^{2} / {2}^{2} = 1}$

#### Explanation:

We are to find out the Standard from of an ellipse

$\text{with vertices at } \left(1 , - 1\right) \mathmr{and} \left(7 , - 1\right)$

$\text{and foci at } \left(4 + \sqrt{5} , - 1\right) \mathmr{and} \left(4 - \sqrt{5} , - 1\right)$

$\text{Center"->(4,-1)" and semi major axis} \left(a\right) = \frac{7 - 1}{2} = 3$

Now if e be eccentricity and b be the semi minor axis of the ellipse then ${e}^{2} = \frac{{a}^{2} - {b}^{2}}{a} ^ 2 \implies {a}^{2} {e}^{2} = {a}^{2} - {b}^{2}$

$\text{The distance of focus from center} = a e = 4 + \sqrt{5} - 4 = \sqrt{5}$

$\implies {a}^{2} {e}^{2} = 5 \implies {a}^{2} - {b}^{2} = 5$

$\implies {3}^{2} - {b}^{2} = 5$

$\implies {b}^{2} = 9 - 5 = 4 \implies b = 2$

So the equation of the ellipse having center$\left(4 , - 1\right)$ and semi major axis $a = 3$ and semi minor axis (b=2)

$\textcolor{red}{{\left(x - 4\right)}^{2} / {3}^{2} + {\left(y + 1\right)}^{2} / {2}^{2} = 1}$