Question

determine the equation in standard form of the ellipse that satisfies the given conditions
center at (1, 3); one vertex at (1, -2); one focus at (1, 0)
question help: video ebook

Explanation:

Step1: Determine the major axis direction

Since the center \((h,k)=(1,3)\), vertex \((1, - 2)\) and focus \((1,0)\) have the same \(x\) - coordinate, the major axis is vertical. The standard form of the ellipse equation is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (\(a>b>0\))

Step2: Calculate \(a\) and \(c\)

The distance from the center \((h,k)\) to a vertex is \(a\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\), for the center \((1,3)\) and vertex \((1,-2)\), \(a=\vert3-(-2)\vert = 5\)
The distance from the center \((h,k)\) to a focus is \(c\). For the center \((1,3)\) and focus \((1,0)\), \(c=\vert3 - 0\vert=3\)

Step3: Calculate \(b^{2}\)

We know the relationship \(c^{2}=a^{2}-b^{2}\), so \(b^{2}=a^{2}-c^{2}\)
Substitute \(a = 5\) and \(c = 3\) into the formula: \(b^{2}=25 - 9=16\)

Step4: Write the equation of the ellipse

Substitute \(h = 1,k = 3,a^{2}=25,b^{2}=16\) into the standard form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\)
We get \(\frac{(x - 1)^{2}}{16}+\frac{(y - 3)^{2}}{25}=1\)

Answer:

\(\frac{(x - 1)^{2}}{16}+\frac{(y - 3)^{2}}{25}=1\)