If $l, m$ are variable real number such that $5 l^{2} + 6 m^{2} - 4 l m + 3 l = 0,$ then variable line $l x + m y = 1$ always touches a fixed parabola , whose axis is parallel to $x - a x i s$

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Vertex of the parabola is $(- \frac{5}{3}, \frac{4}{3})$

Focus of the parabola is $(\frac{1}{3}, \frac{4}{3})$

Di rectrix of the parabola is $3 x + 11 = 0$

Directrix of the parabola is $x + 1 = 0$

answer is A, B, C.

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Detailed Solution

Let EQ of parabola is $(y - k^{2}) = 4 a (x - h);$
Equation of tangent is $y - k = m_{1} (x - h) + a l m_{1}$ Comparing with given tgt $y = - \frac{l}{m} x + \frac{1}{n} .$
And eliminate $m_{1}$ we get $h l^{2} - a m^{2} + k l m - 1$
Now comparing with
$(\frac{- 5}{3}) l^{2} - 2 m^{2} + \frac{4}{3} l m - l = 0$
$\Rightarrow h = \frac{- 5}{3}, a = 2, k = \frac{4}{3}$
Vertex $v (h, k) = (\frac{- 5}{3}, \frac{4}{3})$
Focus $= s (\frac{1}{3}, \frac{4}{3})$
Equation of diretrix is $3 x + 11 = 0$