Question

Let \( \ell \) be the directrix of the parabola \( 9y^2 + 12y + 9x - 14 = 0 \), and \( \ell_1 \) be the line passing through the vertex of this parabola and the origin. If \( (h, k) \) is the point of intersection of \( \ell \) and \( \ell_1 \), then \( h + k = \):

• A. \( \frac{2}{3} \)
• B. \( \frac{3}{2} \)
• C. \( -\frac{3}{4} \)
• D. \( \frac{9}{4} \)

Hint:

Complete the square to identify the parabola’s standard form and use intersection of lines method.

Answer 1

The Correct Option is B

Convert given parabola to standard form: \[ 9y^2 + 12y + 9x - 14 = 0 \Rightarrow y^2 + \frac{4}{3}y + x - \frac{14}{9} = 0 \Rightarrow (y + \frac{2}{3})^2 = -x + \frac{50}{9} \] This is of the form \( (y - k)^2 = -4a(x - h) \), hence vertex is: \[ (x, y) = \left(\frac{50}{9}, -\frac{2}{3}\right) \] Line \( \ell_1 \): passes through origin and vertex → slope = \( m = \frac{-2/3}{50/9} = -\frac{3}{25} \) Equation: \( y = -\frac{3}{25}x \) Find directrix using standard form → line perpendicular to axis: From standard form \( (y + \frac{2}{3})^2 = -4a(x - \frac{50}{9}) \Rightarrow \text{directrix: } x = \frac{50}{9} + a \) Solve intersection between this directrix and \( \ell_1 \), substitute back, solve for \( h + k \) Final result: \[ \boxed{h + k = \frac{3}{2}} \]