Question

$\frac{x^2}{81} + \frac{y^2}{64} = 1$
the major intercepts are at
(0, ±9)
(±8, 0)
(±9, 0)

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse centered at the origin is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\), where if \(a>b\), the major axis is along the \(x\)-axis, and the \(x\)-intercepts (major intercepts in this case) are \((\pm a,0)\).

Step2: Identify \(a^{2}\) from the given equation

Given the equation \(\frac{x^{2}}{81}+\frac{y^{2}}{64}=1\), we can see that \(a^{2}=81\).

Step3: Solve for \(a\)

Taking the square root of both sides, we get \(a=\sqrt{81} = 9\).

Step4: Determine the major intercepts

Since the major axis is along the \(x\)-axis (because \(a = 9\) and \(b=\sqrt{64}=8\), and \(a>b\)), the major intercepts are \((\pm a,0)=(\pm 9,0)\).

Answer:

\((\pm9, 0)\) (corresponding to the option with \((\pm9, 0)\))