Question

Let \(Z\) be the point of intersection of the axis and the directrix of the parabola \(4x^2 - 12x + 4y + 5 = 0\). If \(S\) is its focus, then the point which divides \(SZ\) in the ratio \(2:1\) is:

• A. \(\left(\frac{3}{2}, \frac{13}{12}\right)\)
• B. \(\left(1, \frac{13}{12}\right)\)
• C. \(\left(\frac{3}{4}, \frac{13}{4}\right)\)
• D. \(\left(\frac{3}{2}, \frac{13}{4}\right)\)

Hint:

Convert to standard parabola form by completing square. Then use section formula for internal division.

Answer 1

The Correct Option is A

Given parabola: \(4x^2 - 12x + 4y + 5 = 0 \Rightarrow x^2 - 3x + y + \frac{5}{4} = 0\) Complete square: \((x - \frac{3}{2})^2 - \frac{9}{4} + y + \frac{5}{4} = 0 \Rightarrow (x - \frac{3}{2})^2 + y - 1 = 0\) So vertex = \(\left(\frac{3}{2}, 1\right)\), axis is vertical. Standard form: \((x - h)^2 = 4a(y - k)\), comparing gives \(a = \frac{1}{4}\) Focus = \(\left(\frac{3}{2}, 1 + \frac{1}{4}\right) = \left(\frac{3}{2}, \frac{5}{4}\right)\) Directrix: \(y = 1 - \frac{1}{4} = \frac{3}{4}\) So Z = \(\left(\frac{3}{2}, \frac{3}{4}\right)\) Now, point dividing \(SZ\) in ratio 2:1: Use section formula: \[ \left(\frac{2x_2 + x_1}{3}, \frac{2y_2 + y_1}{3}\right) = \left(\frac{2 \cdot \frac{3}{2} + \frac{3}{2}}{3}, \frac{2 \cdot \frac{3}{4} + \frac{5}{4}}{3}\right) = \left(\frac{6 + 3}{4.5}, \frac{6 + 5}{12}\right) = \left(\frac{3}{2}, \frac{13}{12}\right) \]