Question

If $y = 2 x - 3$ is a tangent to the parabola $y^2 = 4a \left(x\frac{1 }{ 3}\right)$, then ' a ' is equal to :

• A. $\frac{22}{3}$
• B. $-1$
• C. $\frac{14}{3}$
• D. $\frac{-14}{3}$

Answer 1

Answer: (D)

$y = 2x -3 $ and$ y^{2}= 4a\left(x-\frac{1}{3}\right) $
$ \left(2x-3\right)^{2 }= 4a\left(x-\frac{1}{3}\right) \Rightarrow4x^{2}+ 9 - 12 x= 4ax -\frac{4a}{3}\Rightarrow4x^{2}- 4\left(3+a\right)x+9+\frac{4a}{3}$
$ D=0 ; 16\left(3+a^{2}\right)-16\left(9+\frac{4a}{3}\right)=0\Rightarrow9 + a^{2}+6a = 9 +\frac{4a}{3} $
$\Rightarrow a^{2 }+\frac{4a}{3}=0 \Rightarrow a=0 or a = \frac{14}{3}$