Question

7.an ellipse with minor axis from (–2, 1) to (–2, 7) and one focus at (2, 4)write it in standard form

Answer 1

Answer

$\frac{(x+2)^{2}}{25}+\frac{(y-4)^{2}}{9}=1$

Step-by-step explanation

In this case, the minor axis has the coordinates (-2, 1) and (-2, 7), then (referring to the above graph),

h = -2

Given that one focus is placed at (2, 4), then

k = 4

From the point (-2, 7) we can deduce that:

k+b = 7

4+b = 7

b = 7-4

b = 3

From the point (2, 4) we can deduce that:

h+c = 2

-2+c = 2

c = 2+2

c = 4

The relationship between the constants a, b, and c is:

$\begin{gathered} c^2=a^2-b^2 \\ 4^2=a^2-3^2 \\ 16=a^2-9 \\ 16+9=a^2 \\ \sqrt{25}=a \\ a=5 \end{gathered}$

Equation of an ellipse

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

Substituting with the values previously found:

$\begin{gathered} \frac{(x-(-2))^2}{5^2}+\frac{(y-4)^2}{3^2}=1 \\ \frac{(x+2)^2}{25}+\frac{(y-4)^2}{9}=1 \end{gathered}$