Question

an ellipse has a center at the origin, a vertex along the major axis at (13, 0), and a focus at (12, 0). What is the equation of the ellipse?

Answer 1

Hi, as the elipce has a center at the origion and tha major distance is the axis

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

But, 

c = 12        Distance of focus

a = 13   ->  Major distance

b = ?  -> 

But, 

$\\ b^2 + c^2 = a^2 \\ \\ b^2 + 12^2 = 13^2 \\ \\ b^2 + 144 = 169 \\ \\ b^2 = 169 - 144 \\ \\ b^2 = 25 \\ \\ b = \sqrt{5} \\ \\ b = 5$

Then,


$\\ \frac{x^2}{(13)^2} + \frac{y^2}{(5)^2} = 1 \\ \\ or \\ \\ \frac{x^2}{169} + \frac{y^2}{25} = 1$

Answer 2

Answer:  The equation of the ellipse is

$\dfrac{x^2}{169}+\dfrac{y^2}{25}=1.$

Step-by-step explanation:  We are given to find the equation of an ellipse that has a center at the origin, a vertex along the major axis at (13, 0) and a focus at (12, 0).

Since the focus lies on the X-axis, so the standard equation of an ellipse with major axis as X-axis is given by

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

According to given information, we have

a vertex along the major axis, (a, 0) = (13, 0)  

⇒ length of the major axis, a = 13,

co-ordinates of focus, (c, 0) = (12, 0)

⇒ c = 12.

We know that

$c^2=a^2-b^2.$

Therefore,

$12^2=13^2-b^2\\\\\Rightarrow 144=169-b^2\\\\\Rightarrow b^2=169-144\\\\\Rightarrow b^2=25\\\\\Rightarrow b=5.$

Thus, from equation (i), we get

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\\\\\\\Rightarrow \dfrac{x^2}{13^2}+\dfrac{y^2}{5^2}=1\\\\\\\Rightarrow \dfrac{x^2}{169}+\dfrac{y^2}{25}=1.$

The equation of the ellipse is

$\dfrac{x^2}{169}+\dfrac{y^2}{25}=1.$