# Question a7bcf

Jan 28, 2018

The equation of ellipse is ${\left(x + y - 1\right)}^{2} / 8 + {\left(y - x\right)}^{2} / 7 = 1$

#### Explanation:

Focii of the ellipse are at $f 1 \left(0 , 0\right) \mathmr{and} f 2 \left(1 , 1\right)$

Distance between them f1f2= sqrt ((x_1-x_2)^2+(y_1-y_2)^2#

or $f 1 f 2 = \sqrt{{\left(0 - 1\right)}^{2} + {\left(0 - 1\right)}^{2}} = \sqrt{2}$

Distance of the focii from centre is $c = \frac{\sqrt{2}}{2}$

Semi major axis length $a = \frac{4}{2} = 2$ The relation of

$c , a , b$ is ${c}^{2} = {a}^{2} - {b}^{2} \therefore {\left(\frac{\sqrt{2}}{2}\right)}^{2} = {2}^{2} - {b}^{2}$ or

${b}^{2} = 4 - \frac{1}{2} \mathmr{and} b = \sqrt{3.5}$. The center is $\left(h = \frac{1}{2} , k = \frac{1}{2}\right)$

The ellipse is tilted at an angle of $\theta = {\tan}^{-} 1 \left(\frac{1}{1}\right) = {45}^{0}$

Hence the equation of ellipse is.

${\left\{\left(x - h\right) \cos \theta + \left(y - k\right) \sin \theta\right\}}^{2} / {a}^{2} + {\left\{\left(y - k\right) \cos \theta - \left(x - h\right) \sin \theta\right\}}^{2} / {b}^{2} = 1$ or

${\left\{\left(x - \frac{1}{2}\right) \cos 45 + \left(y - \frac{1}{2}\right) \sin 45\right\}}^{2} / 4 + {\left\{\left(y - \frac{1}{2}\right) \cos 45 - \left(x - \frac{1}{2}\right) \sin 45\right\}}^{2} / 3.5 = 1$ or

${\left\{\frac{x - \frac{1}{2}}{\sqrt{2}} + \frac{y - \frac{1}{2}}{\sqrt{2}}\right\}}^{2} / 4 + {\left\{\frac{y - \frac{1}{2}}{\sqrt{2}} - \frac{x - \frac{1}{2}}{\sqrt{2}}\right\}}^{2} / 3.5 = 1$

${\left\{\frac{2 x - 1}{2 \sqrt{2}} + \frac{2 y - 1}{2 \sqrt{2}}\right\}}^{2} / 4 + {\left\{\frac{2 y - 1}{2 \sqrt{2}} - \frac{2 x - 1}{2 \sqrt{2}}\right\}}^{2} / 3.5 = 1$ or

${\left\{\frac{2 x - 1 + 2 y - 1}{2 \sqrt{2}}\right\}}^{2} / 4 + {\left\{\frac{2 y - 2 x}{2 \sqrt{2}}\right\}}^{2} / 3.5 = 1$ or

${\left(x + y - 1\right)}^{2} / 8 + {\left(y - x\right)}^{2} / 7 = 1$ [Ans]