Question

The length of the latus rectum of the parabola \( y^2-4x-4y+8=0 \) is

• A. 2
• B. 4
• C. 6
• D. 8

Hint:

Standard forms of parabola:
\((y-k)^2 = 4a(x-h)\) (opens right if a>0, left if a<0). Latus rectum = \(|4a|\).
\((x-h)^2 = 4a(y-k)\) (opens up if a>0, down if a<0). Latus rectum = \(|4a|\).
Complete the square to bring the given equation into standard form.
The length of the latus rectum is the absolute value of the coefficient of the linear term (multiplied by 4 if it's in the form \( (y-k)^2 = \text{coeff} \cdot (x-h) \), then latus rectum is |coeff|).

Answer 1

The Correct Option is B

The given equation of the parabola is \( y^2-4x-4y+8=0 \). To find the length of the latus rectum, we need to rewrite this equation in the standard form of a parabola. Since there is a \(y^2\) term and a \(y\) term, but only an \(x\) term (not \(x^2\)), the parabola opens horizontally. Group terms in y: \( (y^2 - 4y) - 4x + 8 = 0 \) Complete the square for the y terms: \(y^2 - 4y = (y-2)^2 - 4\). Substitute this back: \( (y-2)^2 - 4 - 4x + 8 = 0 \) \( (y-2)^2 - 4x + 4 = 0 \) \( (y-2)^2 = 4x - 4 \) \( (y-2)^2 = 4(x - 1) \) This is of the standard form \( (y-k)^2 = 4a(x-h) \), where the vertex is \((h,k)\) and the length of the latus rectum is \(|4a|\). Comparing \( (y-2)^2 = 4(x - 1) \) with \( (y-k)^2 = 4a(x-h) \): Vertex \((h,k) = (1,2)\). \(4a = 4\). The length of the latus rectum is \(|4a|\). Length of latus rectum = \(|4| = 4\). This matches option (b). \[ \boxed{4} \]