Question

an equation of an ellipse is given.
(a) write the equation of the ellipse in standard form.
(b) identify the center, vertices, endpoints of the minor axis, and foci.
express numbers in exact, simplest form.
$x^{2}+25 y^{2}-2 x-24 = 0$

Explanation:

Step1: Complete the square for \(x\) terms

Given equation \(x^{2}+25y^{2}-2x - 24=0\).
Group \(x\) terms: \((x^{2}-2x)+25y^{2}=24\).
Complete the square for \(x\): \((x - 1)^{2}-1+25y^{2}=24\).
So, \((x - 1)^{2}+25y^{2}=25\).
Divide by 25: \(\frac{(x - 1)^{2}}{25}+\frac{y^{2}}{1}=1\).

Step2: Identify center, vertices, endpoints of minor axis and foci

For the ellipse \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) (\(a>b>0\)), here \(h = 1,k = 0,a = 5,b = 1\).
Center \((h,k)=(1,0)\).
Vertices: \((h\pm a,k)\), so \((1\pm5,0)\) i.e. \((6,0)\) and \((-4,0)\).
Endpoints of minor axis: \((h,k\pm b)\), so \((1,0\pm1)\) i.e. \((1,1)\) and \((1,-1)\).
Calculate \(c=\sqrt{a^{2}-b^{2}}=\sqrt{25 - 1}=\sqrt{24}=2\sqrt{6}\).
Foci: \((h\pm c,k)\), so \((1\pm2\sqrt{6},0)\).

Answer:

(a) The standard form of the ellipse is \(\frac{(x - 1)^{2}}{25}+\frac{y^{2}}{1}=1\).
(b) Center: \((1,0)\); Vertices: \((6,0)\) and \((-4,0)\); Endpoints of minor axis: \((1,1)\) and \((1,-1)\); Foci: \((1 + 2\sqrt{6},0)\) and \((1-2\sqrt{6},0)\).