Question

question 9 · 1 point what is the standard form equation of the ellipse that has vertices (±12,0) and foci (±9,0)? select the correct answer below: $\frac{x^{2}}{144}+\frac{y^{2}}{3}=1$ $\frac{x^{2}}{12}+\frac{y^{2}}{3}=1$ $\frac{x^{2}}{12}+\frac{y^{2}}{9}=1$ $\frac{x^{2}}{144}+\frac{y^{2}}{9}=1$ $\frac{x^{2}}{12}+\frac{y^{2}}{63}=1$ $\frac{x^{2}}{144}+\frac{y^{2}}{63}=1$

Explanation:

Step1: Identify the value of $a$

Vertices are $(\pm12,0)$, so $a = 12$ and $a^{2}=144$.

Step2: Identify the value of $c$

Foci are $(\pm9,0)$, so $c = 9$.

Step3: Use the relationship $c^{2}=a^{2}-b^{2}$

$b^{2}=a^{2}-c^{2}$. Substitute $a = 12$ and $c = 9$, we get $b^{2}=144 - 81=63$.

Step4: Write the standard - form equation

The standard - form equation of an ellipse with a horizontal major axis is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$. Substituting $a^{2}=144$ and $b^{2}=63$, we have $\frac{x^{2}}{144}+\frac{y^{2}}{63}=1$.

Answer:

F. $\frac{x^{2}}{144}+\frac{y^{2}}{63}=1$