Question #d2701

Question

Question #d2701

1 Answer

Impact of this question

1933 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-20T23:19:05

$\frac{{(x - 4)}^{2}}{3^{2}} + \frac{{(y + 1)}^{2}}{2^{2}} = 1$ $color(red)((x-4)^2/3^2+(y+1)^2/2^2=1)$

Explanation:

We are to find out the Standard from of an ellipse

$with vertices at (1, - 1) and (7, - 1)$ $"with vertices at "(1, -1) and (7, -1)$

$and foci at (4 + \sqrt{5}, - 1) and (4 - \sqrt{5}, - 1)$ $"and foci at "(4+sqrt5, -1) and (4-sqrt5, -1)$

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

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

$The distance of focus from center = a e = 4 + \sqrt{5} - 4 = \sqrt{5}$ $"The distance of focus from center"=ae=4+sqrt5-4=sqrt5$

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

$\Rightarrow 3^{2} - b^{2} = 5$ $=>3^2-b^2=5$

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

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

$\frac{{(x - 4)}^{2}}{3^{2}} + \frac{{(y + 1)}^{2}}{2^{2}} = 1$ $color(red)((x-4)^2/3^2+(y+1)^2/2^2=1)$

Science

Math

Humanities

... and beyond

Question #d2701

1 Answer

Explanation:

Related questions

Impact of this question