Question

The eccentricity of the curve \[ 2x^2 + y^2 - 8x - 2y + 1 = 0 { is:} \]

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{3}{4} \)

Hint:

To find the eccentricity of an ellipse, use the formula \( e = \sqrt{1 - \frac{b^2}{a^2}} \) and rewrite the equation in standard form.

Answer 1

The Correct Option is B

Step 1: Rewrite the equation of the conic in standard form. First, complete the square for \( x \) and \( y \). For \( x \), the coefficient of \( x \) is -8. Half of -8 is -4, and \( (-4)^2 = 16 \). For \( y \), the coefficient of \( y \) is -2. Half of -2 is -1, and \( (-1)^2 = 1 \). Add and subtract these values inside the equation: \[ 2(x^2 - 4x + 16) + (y^2 - 2y + 1) = 1 - 32 + 2. \] Simplifying: \[ 2(x - 2)^2 + (y - 1)^2 = 1. \] Divide through by 1: \[ \frac{(x - 2)^2}{\frac{1}{2}} + \frac{(y - 1)^2}{1} = 1. \] Step 2: This is the equation of an ellipse in standard form: \[ \frac{(x - 2)^2}{a^2} + \frac{(y - 1)^2}{b^2} = 1, \] where \( a^2 = \frac{1}{2} \) and \( b^2 = 1 \). The eccentricity \( e \) of an ellipse is given by: \[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{\frac{1}{2}}} = \sqrt{1 - 2} = \frac{1}{\sqrt{2}}. \] Thus, the eccentricity is \( \frac{1}{\sqrt{2}} \).