# I have had a hard time with this question, what is the standard form of the equation of the ellipse with the given characteristics? Center: (1, −5); a = 7c; foci: (1, −6), (1, −4)

Apr 26, 2018

${\left(x - 1\right)}^{2} / 48 + {\left(y + 5\right)}^{2} / 49 = 1$

#### Explanation:

The standard form of the equation of an ellipse is

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

where $\left(h , k\right)$ is the center of the ellipsecolor(green)( rarr(1,-5)

$a = 7$ color(green)(rarr(given)

The distance between the two foci is $2 a e$

$\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = 2 a e$

$\sqrt{{\left(1 - 1\right)}^{2} + {\left(- 4 - \left(- 6\right)\right)}^{2}} = 2 a e$

$2 = 2 \times 7 \times e$

$e = \frac{1}{7}$

${b}^{2} = {a}^{2} \left(1 - {e}^{2}\right)$

${b}^{2} = 48$

Now Substitute in the equation's standard form

${\left(x - 1\right)}^{2} / 48 + {\left(y - \left(- 5\right)\right)}^{2} / 49 = 1$

${\left(x - 1\right)}^{2} / 48 + {\left(y + 5\right)}^{2} / 49 = 1$

color(blue)("We use this form as the ellipse is vertical while we would use this form "
color(blue)("if it were horizontal " (x-h)^2/a^2+(y-k)^2/b^2=1

color(green)("And to determine whether it's vertical or horizontal you could use the graph"

graph{(x-1)^2/48+(y+5)^2/49=1 [-14.72, 17.32, -13, 3.02]}

color(green)("Or you could identify it using the given points as in this question the foci are (1,-6) and (1,-4) so the difference between them is in the " y " value so it's vertical"