Question

can someone tell me what are the steps to solving this?

Answer 1

The general equation of an ellipse is

$\dfrac{(x-x_0)^2}{a^2}+\dfrac{(y-y_0)^2}{b^2} = 1$

where:

• $x_0$ is the x coordinate of the center
• $y_0$ is the y coordinate of the center
• $a$ is half the length of the major axis
• $b$ is half the length of the minor axis

So, in your case, we have $(x_0,y_0)=(2,3)$, and the equation will look like this:

$\dfrac{(x-2)^2}{a^2}+\dfrac{(y-3)^2}{b^2} = 1$

Moreover, we know that the major axis is horizontal (i.e. it involves the x coordinate). Its length is 8, which implies $a=4$

Similarly, the minor axis has length 4, which implies $b=2$

So, the complete equation is

$\dfrac{(x-2)^2}{16}+\dfrac{(y-3)^2}{4} = 1$