Question

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

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

Answer 1

The correct option is D $\frac{-14}{3}$

Solving $y=2x-3$ and $y^2=4a(x-\frac{1}{3})$, we have

$(2x-3{)}^2=4a(x-\frac{1}{3})$

$\Rightarrow 4x^2+9-12x=4ax-\frac{4a}{3}$

$\Rightarrow 4x^2-4(3+a)x+9+\frac{4a}{3}$

$\Rightarrow 4x^2-4(3+a)x+9+\frac{4a}{3}=0$

This equation must have equal roots $\Rightarrow D=0$

$\Rightarrow 16(3+a{)}^2-16(9+\frac{4a}{3})=0$

$\Rightarrow 9+a^2+6a=9+\frac{4a}{3}$

$\Rightarrow a^2+\frac{14a}{3}=0$

$\Rightarrow a=0$ or $a=-\frac{14}{3}$