Question

$\frac{x^2}{9} + \frac{y^2}{49} = 1$
the distance from the center to the major intercept is
○ 9
○ 49
○ 7

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (when \(a>b>0\)), where the center is at \((0,0)\), the major axis is along the \(y\)-axis, the length of the semi - major axis is \(a\), and the length of the semi - minor axis is \(b\).

For the given ellipse equation \(\frac{x^{2}}{9}+\frac{y^{2}}{49}=1\), we can rewrite it as \(\frac{x^{2}}{3^{2}}+\frac{y^{2}}{7^{2}} = 1\). Here, \(a = 7\) and \(b=3\) (since \(a>b\) and the major axis is along the \(y\) - axis).

Step2: Determine the distance from the center to the major intercept

The center of the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) is at the origin \((0,0)\). The major intercepts (the endpoints of the major axis) are at \((0,\pm a)\). The distance from the center \((0,0)\) to a major intercept (e.g., \((0,a)\)) is given by the value of \(a\). Since \(a = 7\), the distance from the center to the major intercept is \(7\).

Answer:

7