Question

If the line $y=kx$ touches the parabola $y={(x-1)}^2$, then the values of $k$ are

• A. $2,-2$
• B. $0,4$
• C. $0,-2$
• D. $0,2$
• E. $0,-4$

Answer 1

The correct option is E $0,-4$

Put $y=kx$ in $y={(x-1)}^2$, we get
$kx={(x-1)}^2$
$\Rightarrow x^2-(k+2)x+1=0$ ....(i)
For line $y=kx$ to be tangent, discriminant of the equation (i) must be zero.
Therefore, ${(k+2)}^2-4=0$
$\Rightarrow {(k+2)}^2=4$
$\Rightarrow k+2=\pm 2$
$\Rightarrow k=\pm 2-2$

$\Rightarrow k=0,-4$