Question

What is an equation in standard form of an ellipse centered at the origin with vertex (-6,0) and co-vertex (0,4)?

Answer 1

$\frac{x^{2} }{36}$ + $\frac{ y^{2}}{16}$ = 1

This is because the standard form of an ellipse is 

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

where a is the vertex and b is the co-vertex. So when we stick their respective x and y values in and then square them, you're left with the answer above. 

Answer 2

Answer:

$\frac{x^{2} }{36}+\frac{y^{2} }{16}=1$

Step-by-step explanation:

As we know standard equation of n ellipse is $\frac{(x-0)^{2} }{(-6)^{2}}+\frac{(y-0)^{2} }{(4)^{2} }=1$

In the given equation (h, k) is the center, a is the vertex and b is the co-vertex.

Here vertex is (-6, 0) and co-vertex is (0, 4)

Therefore, length of a = -6, b = 4 and origin is (0, 0)

Now the equation of the ellipse will be

$\frac{(x-0)^{2} }{(-6)^{2}}+\frac{(y-0)^{2} }{(4)^{2} }=1$

$\frac{x^{2} }{6^{2}}+\frac{y^{2} }{4^{2} }=1$

$\frac{x^{2} }{36}+\frac{y^{2} }{16}=1$

Therefore, the equation of the ellipse will be $\frac{x^{2} }{36}+\frac{y^{2} }{16}=1$