Question

The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}+\frac{4(y-1)^2}{45}=1\) is equal to

• A. \(\frac{3}{\sqrt{5}}\)
• B. \(\frac{5}{3}\)
• C. \(\frac{5}{\sqrt{3}}\)
• D. \(\frac{5}{2}\)
• E. \(\frac{3}{2}\)

Answer 1

Answer: (E)

The standard form of a hyperbola equation is \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1, \] where \( (h, k) \) is the center, \( a \) is the distance from the center to the vertices along the x-axis, and \( b \) is the distance from the center to the vertices along the y-axis. The eccentricity \( e \) is given by the formula: \[ e = \sqrt{1 + \frac{b^2}{a^2}}. \] For the given hyperbola \( \frac{(x - 3)^2}{9} - \frac{4(y - 1)^2}{45} = 1 \), we have \( a^2 = 9 \) and \( b^2 = \frac{45}{4} \). Now, substitute these values into the eccentricity formula: \[ e = \sqrt{1 + \frac{\frac{45}{4}}{9}} = \sqrt{1 + \frac{45}{36}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}. \]

The correct option is (E) : \(\frac{3}{2}\)

Answer 2

The given equation of the hyperbola is \(\frac{(x-3)^2}{9} - \frac{4(y-1)^2}{45} = 1\). We can rewrite this as \(\frac{(x-3)^2}{9} - \frac{(y-1)^2}{\frac{45}{4}} = 1\).

Comparing this with the standard equation of a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), we have \(a^2 = 9\) and \(b^2 = \frac{45}{4}\).

So, \(a = \sqrt{9} = 3\) and \(b = \sqrt{\frac{45}{4}} = \frac{\sqrt{45}}{2} = \frac{3\sqrt{5}}{2}\).

The eccentricity \(e\) of a hyperbola is given by the formula \(c^2 = a^2 + b^2\), where \(c\) is related to eccentricity by \(e = \frac{c}{a}\). So, we need to find \(c\) first.

\(c^2 = a^2 + b^2 = 9 + \frac{45}{4} = \frac{36 + 45}{4} = \frac{81}{4}\)

Therefore, \(c = \sqrt{\frac{81}{4}} = \frac{9}{2}\).

Now we can find the eccentricity:

\(e = \frac{c}{a} = \frac{\frac{9}{2}}{3} = \frac{9}{2} \cdot \frac{1}{3} = \frac{3}{2}\)

Therefore, the eccentricity of the hyperbola is \(\frac{3}{2}\).

There is an error in the question. Should be minus sign between the two terms:

The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}-\frac{4(y-1)^2}{45}=1\) is equal to Comparing this with the standard equation of a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), we have \(a^2 = 9\) and \(b^2 = \frac{45}{4}\). So, \(a = \sqrt{9} = 3\) and \(b = \sqrt{\frac{45}{4}} = \frac{\sqrt{45}}{2} = \frac{3\sqrt{5}}{2}\). 

The eccentricity \(e\) of a hyperbola is given by the formula \(c^2 = a^2 + b^2\), where \(c\) is related to eccentricity by \(e = \frac{c}{a}\). 

So, we need to find \(c\) first. \(c^2 = a^2 + b^2 = 9 + \frac{45}{4} = \frac{36 + 45}{4} = \frac{81}{4}\) Therefore, \(c = \sqrt{\frac{81}{4}} = \frac{9}{2}\). 

Now we can find the eccentricity: \(e = \frac{c}{a} = \frac{\frac{9}{2}}{3} = \frac{9}{2} \cdot \frac{1}{3} = \frac{3}{2}\) Therefore, the eccentricity of the hyperbola is \(\frac{3}{2}\).