Question

graph the ellipse equation.
5.1) (\frac{x^2}{49} + \frac{y^2}{25} = 1)

• is the ellipse horizontal or vertical?
• (a = )
• (b = )
• (h = )
• (k = )
• vertices (= (h - a, k)(h + a, k) = (quad)(quad) = (quad)(quad)
• co - vertices (= (h, k - b)(h, k + b) = (quad)(quad) = (quad)(quad)

Explanation:

Step1: Recall Ellipse Standard Form

The standard form of an ellipse centered at \((h, k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (horizontal major axis) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (vertical major axis), where \(a > b>0\). For \(\frac{x^2}{49}+\frac{y^2}{25}=1\), rewrite as \(\frac{(x - 0)^2}{7^2}+\frac{(y - 0)^2}{5^2}=1\). So \(h = 0\), \(k = 0\), \(a = 7\) (since \(49=7^2\)), \(b = 5\) (since \(25 = 5^2\)). Since \(a^2\) is under \(x^2\), the major axis is horizontal.

Step2: Find Vertices

Vertices for horizontal ellipse: \((h\pm a,k)\). Substitute \(h = 0\), \(k = 0\), \(a = 7\): \((0 - 7,0)=(-7,0)\) and \((0 + 7,0)=(7,0)\).

Step3: Find Co - vertices

Co - vertices for horizontal ellipse: \((h,k\pm b)\). Substitute \(h = 0\), \(k = 0\), \(b = 5\): \((0,0 - 5)=(0,-5)\) and \((0,0 + 5)=(0,5)\).

Answer:

• Is the ellipse horizontal or vertical? Horizontal
• \(a=\boldsymbol{7}\), \(b=\boldsymbol{5}\), \(h=\boldsymbol{0}\), \(k=\boldsymbol{0}\)
• Vertices: \((-7,0)\), \((7,0)\)
• Co - vertices: \((0,-5)\), \((0,5)\)