Question

The eccentricity of the hyperbola $25x^{2}-36y^{2}-50x-72y-911=0$, is

• A. $\frac{\sqrt{61}}{6}$
• B. $\frac{\sqrt{65}}{6}$
• C. $\frac{\sqrt{61}}{4}$
• D. $\frac{\sqrt{59}}{6}$
• E. $\frac{\sqrt{71}}{6}$

Hint:

Math Tip: To quickly find the squared terms for completing the square, take half of the linear coefficient and square it. Always remember to multiply that squared value by the factored coefficient outside the bracket before adding it to the right side of the equation!

Answer 1

The Correct Option is A

Concept:
Coordinate Geometry - Hyperbola and Eccentricity.
The standard equation of a hyperbola is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$.
Its eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Step 1: Group the $x$ and $y$ terms together.
Given equation: $25x^2 - 36y^2 - 50x - 72y - 911 = 0$
Rearrange the terms: $$ (25x^2 - 50x) - (36y^2 + 72y) = 911 $$
Step 2: Factor out the coefficients of the squared terms.
$$ 25(x^2 - 2x) - 36(y^2 + 2y) = 911 $$
Step 3: Complete the square for both groups.
Add the necessary constants inside the parentheses and balance the equation on the right side: $$ 25(x^2 - 2x + 1) - 36(y^2 + 2y + 1) = 911 + 25(1) - 36(1) $$ $$ 25(x - 1)^2 - 36(y + 1)^2 = 911 + 25 - 36 $$ $$ 25(x - 1)^2 - 36(y + 1)^2 = 900 $$
Step 4: Convert to standard form.
Divide the entire equation by 900: $$ \frac{25(x - 1)^2}{900} - \frac{36(y + 1)^2}{900} = 1 $$ $$ \frac{(x - 1)^2}{36} - \frac{(y + 1)^2}{25} = 1 $$
Step 5: Identify $a^2$ and $b^2$.
Comparing with the standard form:

• $a^2 = 36$
• $b^2 = 25$

Step 6: Calculate the eccentricity.
Use the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$: $$ e = \sqrt{1 + \frac{25}{36}} $$ $$ e = \sqrt{\frac{36 + 25}{36}} $$ $$ e = \sqrt{\frac{61}{36}} = \frac{\sqrt{61}}{6} $$