Question

The eccentricity of the conic \(x^2 - 4x + 4y^2 = 12\) is

• A. \(\frac{\sqrt{3}}{2}\)
• B. \(\frac{2}{\sqrt{3}}\)
• C. \(\sqrt{3}\)
• D. None of these

Hint:

For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a>b\), \(e = \sqrt{1 - b^2/a^2}\).

Answer 1

The Correct Option is A

To determine the eccentricity of the conic given by the equation \( x^2 - 4x + 4y^2 = 12 \), we first need to identify the type of conic and express it in standard form. 

1. Start with the given equation: \(x^2 - 4x + 4y^2 = 12\).
2. Complete the square for the \(x\) term:
   • \(x^2 - 4x = (x - 2)^2 - 4\)
3. Substitute back into the equation:
   • \((x - 2)^2 - 4 + 4y^2 = 12\)
   • \((x - 2)^2 + 4y^2 = 16\)
4. Divide the entire equation by 16 to convert it into standard form:
   • \(\frac{(x - 2)^2}{16} + \frac{y^2}{4} = 1\)
5. This equation represents an ellipse of the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\):
   • Here, \(a = 4\), \(b = 2\), and \(h = 2, k = 0\).
   • For an ellipse, the eccentricity \(e\) is given by \(e = \sqrt{1 - \frac{b^2}{a^2}}\).
6. Substitute \(a\) and \(b\) into the eccentricity formula:
   • \(e = \sqrt{1 - \frac{4}{16}}\)
   • \(e = \sqrt{1 - \frac{1}{4}}\)
   • \(e = \sqrt{\frac{3}{4}}\)
   • \(e = \frac{\sqrt{3}}{2}\)
7. Therefore, the eccentricity of the given conic is \(\frac{\sqrt{3}}{2}\).