Question

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

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

Answer 1

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

$\begin{array}{c}y=2x-3is\tan gent{y}^2=4x\left(x-\frac{1}{3}\right)\\ y=2x-3\\ {\left(2x-3\right)}^2=4a\left(x-\frac{1}{3}\right)\\ \Rightarrow 4x^2+9-12x=4ax-\frac{4}{3}a\\ \Rightarrow 4x^2-4\left(3+a\right)x+9+\frac{4x}{3}=0\\ equalroots\\ D=0\\ 16{\left(3+a\right)}^2-4\times 4\times \left(9+\frac{4x}{3}\right)=0\\ \Rightarrow 9+a^2+6a-9-\frac{4a}{3}=0\\ \Rightarrow a^2+6a-\frac{4a}{3}=0\\ \Rightarrow \frac{3a^2+18a-4a}{3}=0\\ \Rightarrow 3a^2+14a=0\\ \therefore a=0,a=-\frac{14}{3}\end{array}$

Hence, option $D$ is the required answer.