Question

The one end of the latus rectum of the parabola \( y^2 - 4x - 2y - 3 = 0 \) is at

• A. \( (0,-1) \)
• B. \( (0,1) \)
• C. \( (0,-3) \)
• D. \( (3,0) \)
• E. \( (0,2) \)

Hint:

After converting to standard form, directly use \( (h+a, k \pm 2a) \) for latus rectum endpoints.

Answer 1

The Correct Option is A

Concept:
• Convert given equation into standard form of parabola.
• Standard form: \( (y-k)^2 = 4a(x-h) \)
• Focus: \( (h+a, k) \)
• Endpoints of latus rectum: \( (h+a, k \pm 2a) \)

Step 1: Rearrange equation
\[ y^2 - 2y - 4x - 3 = 0 \] Group terms: \[ y^2 - 2y = 4x + 3 \]

Step 2: Complete the square
\[ y^2 - 2y + 1 = 4x + 3 + 1 \] \[ (y - 1)^2 = 4x + 4 \] \[ (y - 1)^2 = 4(x + 1) \]

Step 3: Compare with standard form
\[ (y - k)^2 = 4a(x - h) \] So, \[ h = -1,\quad k = 1,\quad 4a = 4 \Rightarrow a = 1 \]

Step 4: Find focus
\[ \text{Focus} = (h+a, k) = (-1+1, 1) = (0,1) \]

Step 5: Find endpoints of latus rectum
\[ (h+a, k \pm 2a) = (0, 1 \pm 2) \] \[ = (0,3) \quad \text{and} \quad (0,-1) \]

Step 6: Select required answer
One end is: \[ \boxed{(0,-1)} \]