Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $\frac{-14}{3}$

Answer 1

Answer: (C), (D)

$0$
$\frac{-14}{3}$

For given two values of $a$ in option the given straight line never be a tangent to the given parabola.

Argument :-

If possible let $y=2x-3$ be tangent to the parabola $y^2=4a(x-\frac{1}{3})$ at $(h,k)$.

To find the touching point we have the following equations,

$k=2h-3$........(1) and

$k^2=4a(h-\frac{1}{3})$........(2)

Using $\left(1\right)$ in $\left(2\right)$, we get

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

or, $4h^2-12h+9=4ah-\frac{4a}{3}$

or, $4h^2-(12+4a)h+9+\frac{4a}{3}=0$

$\Rightarrow h=\frac{(12+4a)\pm \sqrt{(12+4a{)}^2-16(9+\frac{4a}{3})}}{2\times 4}$

$\Rightarrow h=\frac{(3+a)\pm \sqrt{(3+a{)}^2-(9+\frac{4a}{3})}}{2}$

$\Rightarrow h=\frac{(3+a)\pm \sqrt{a^2+\frac{14a}{3}}}{2}$

$\Rightarrow h=\frac{(3+a)\pm \sqrt{a(a+\frac{14}{3})}}{2}$

Only for $a=0$ and $a=-\frac{14}{3}$, the given st.line will be a tangent to the parabola at $(h,k)$.