Line $x+ky+k^2=0$ is tangent to curve $y^2=4x$ at point $P$. Find the coordinates of $P$ in terms of $k$.

Question

The line $x + ky +k^2 = 0$, where $k$ is a constant, is tangent to curve $y^2 = 4x$ at the point $P$. Find the coordinates of $P$ in terms of $k$.

I solved for $y$ getting $2\sqrt x$ and then substituted this in the equation for tangent to get $x + 2k\sqrt x +k^2 = 0$. Then I set $u = \sqrt x$ to obtain a quadratic equation in terms of $u$ and $k$. Next, I solved for $u$ in terms of $k$ and finally got $x = k^2$ and $y$ as $2k$.

However, the answer in the book says $y$ is $-2k$. How does that make sense?

Answer 1

We are told that $$x=-ky-k^2$$ is tangential to $$x=\frac14y^2$$

We have $$\frac{dx}{dy}=\frac24y$$

$$\frac{2}{4}y=-k$$

Hence $y=-2k$ and $x=\frac14(-2k)^2=k^2$.

Answer 2

The line x+ky+k²=0 is tangent to y²=4x , which is a parabola .

Substituting x=y²/4 , in the equation of the given line , we get

y²+4ky+k²=0 .

Now as the line is a tangent , there is only one point of intersection , hence this quadratic equation has equal roots , therefore the determinant (b²-4ac) = 0 . Now by quadratic formula . Roots = (-b+√D)/2a or (-b-√D)/2a , but as D=0 , both roots are -b/2a , which is equal to -4k/2= -2k . Therefore , y=-2k.