Question

If the line $ky-2x-k^2+2h=0\&$ parabola $x^2=4y$ touches each other, then

• A. $\frac{k}{2}(k^2+2h)=2$
• B. $\frac{k}{4}(k^2+2h)=4$
• C. $\frac{k}{4}(-k^2+2h)=2$
• D. $\frac{k}{4}(-k^2+2h)=4$

Answer 1

The correct option is D

$\frac{k}{4}(-k^2+2h)=4$

Solving these two equations

$ky-2x-k^2+2h=0$ - - - - - - - - - (1) & parabola $x^2=4y-------\left(2\right)$ is

substituting $y=\frac{x^2}{4}$ in equation (1)

$k.\frac{x^2}{4}-2x-k^2+2h=0$

$\frac{k}{4}{x}^2-2x-k^2+2h$

since ,it touches to parabola ;there should be only one root

discriminant (D)=0

$b^2-4ac=0$

$4-4\times \left(\frac{k}{4}\right).(-k^2+2h)=0$

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

$\frac{k}{4}(-k^2+2h)=4$