Question

If $y=2x-1$ is a tangent to the parabola $y^2=4a(x+1)$, then $a$ is equal to:

• A. $6$
• B. $-6$
• C. $-3+2\sqrt{2}$
• D. $-3-2\sqrt{2}$

Answer 1

The correct option is B $-6$

If the line $y=2x-1$ is a tangent to the parabola $y^2=4a(x+1)$, then it will touch the parabola at a single point.

To find the point of contact, substitute $y=2x-1$ in the parabola equation.

$(2x-1{)}^2=4a(x+1)$

$4x^2+1-4x=4ax+4a$

$4x^2-(4a+4)x-4a+1=0$

If the above equation has only one solution for $x$, then

$b^2-4ac=0$

$(4a+4{)}^2-4(4\left)\right(1-4a)=0$

$16a^2+96a=0$

$a^2+6a=0$

$a=0,-6$

$a$ cannot be equal to zero, because then parabola will not exist.

therefore $a=-6$