Question

find the ends of the major axis and foci of this ellipse. $\frac{x^{2}}{81}+\frac{y^{2}}{16}=1$ major axis : $(pm?,0)$ foci : $(pmsqrt{},0)$

Explanation:

Step1: Identify $a^2$ and $b^2$

The standard form of an ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$. Here, $a^{2}=81$, $b^{2}=16$.

Step2: Find the ends of the major - axis

For an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$ with $a>b$, the major - axis lies on the x - axis and its ends are $(\pm a,0)$. Since $a^{2}=81$, then $a = 9$. So the ends of the major axis are $(\pm9,0)$.

Step3: Calculate the value of $c$

The relationship for an ellipse is $c^{2}=a^{2}-b^{2}$. Substitute $a^{2}=81$ and $b^{2}=16$ into the formula: $c^{2}=81 - 16=65$. So $c=\sqrt{65}$, and the foci are $(\pm\sqrt{65},0)$.

Answer:

Major axis: $(\pm9,0)$
Foci: $(\pm\sqrt{65},0)$