Question

The center of an ellipse is (-9,3). one focus is (-7,3). The major axis is 24 units long. What is the equation of the ellipse in standard form?
I got as far as ((x+9)^2)/24 + ((y-3)^2)/b^2 =1

But I don't know how to get b^2 (length of the minor axis)​

Answer 1

Answer:

$\frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1$

Step-by-step explanation:

The standard equation of the ellipse is

$\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1\;\cdots(i)$

where, $(\alpha, \beta)$ is the center and a,b are semi axes of the ellipse along x-axis and y-axis respectively.

Given that, $(\alpha, \beta)=(-9,3)$ and major axis, $2a=24$.

So, semi major axis, $a=24/2=12$.

Now, focus of the ellipse is (-7,3).

Let $e$ be the eccentricity of the ellipse,

So, $e=\frac c a$

where $c$ is the distance between the center and focus of the ellipse.

By using the distance formula,

$c=\sqrt{(-9-(-7))^2+(3-3)^2}=2$

$\Rightarrow e=\frac {2}{12}=\frac{1}{6}\;\cdots(ii)$

Again, the relationship among $a,b$ and $e$ is

$e=\sqrt{1-\frac{b^2}{a^2}$

$\Rightarrow \frac{1}{6}=\sqrt{1-\frac{b^2}{(12)^2}$ [from equation (ii)]

$\Rightarrow \frac{1}{36}=\sqrt{1-\frac{b^2}{144}$ [squaring on both the sides]

$\Rightarrow \frac{b^2}{144}=1-\frac{1}{36}$

$\Rightarrow b^2=\frac{35}{36}\times144=140$

So, the value of square of semi-minir axis, $b^2=140$.

Hence, from equation (i), the equation of required ellipse is standard form is

$\frac{(x-(-9))^2}{144}+\frac{(y-3)^2}{140}=1$

$\Rightarrow \frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1$