Question

for ( 10 x ^ { 2 } - y ^ { 2 } + 2 y - 11 = 0 ).
(a) write the equation of the hyperbola in standard form.
(b) identify the center, vertices, and foci.
give exact answers. express numbers in simplest form using integers, fractions, or radicals.

Explanation:

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

Given \(10x^{2}-y^{2}+2y - 11=0\).
Group \(y\) terms: \(10x^{2}-(y^{2}-2y)=11\).
Complete the square for \(y^{2}-2y\): \(y^{2}-2y=(y - 1)^{2}-1\).
So the equation becomes \(10x^{2}-[(y - 1)^{2}-1]=11\), which simplifies to \(10x^{2}-(y - 1)^{2}=10\).
Divide by \(10\) to get the standard form \(\frac{x^{2}}{1}-\frac{(y - 1)^{2}}{10}=1\).

Step2: Identify center, vertices and foci

For a hyperbola of the form \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\), here \(h = 0,k = 1,a^{2}=1,b^{2}=10\).

• Center: \((h,k)=(0,1)\)
• Vertices: Since \(a = 1\) and the hyperbola opens left - right (because \(x\) term is positive), vertices are \((h\pm a,k)=( \pm1,1)\)
• Foci: Calculate \(c\) using \(c^{2}=a^{2}+b^{2}\). So \(c^{2}=1 + 10=11\), \(c=\sqrt{11}\). Foci are \((h\pm c,k)=( \pm\sqrt{11},1)\)

Answer:

(a) The standard form of the hyperbola is \(\frac{x^{2}}{1}-\frac{(y - 1)^{2}}{10}=1\)
(b) Center: \((0,1)\), Vertices: \((1,1)\) and \((- 1,1)\), Foci: \((\sqrt{11},1)\) and \((-\sqrt{11},1)\)