The line $y=\frac{x}{k}+k$ ,where k is a constant, is a tangent to the curve $4y=x^2$ at the point $P$. I have to find the value of k and the coordinates of $P$
I'm stuck and I don't what I'm doing wrong.To find K , I know I have to substitute , so I get
$4(\frac{x}{k}+k) =x^2$
$\frac{4x}{k}+k =x^2$
$4x+4k^2=x^2$
Then I use the discriminant $b^2 -4ac$ which gives
16 + 16k^2
$k^2 =-1$
I can't find $k$ because $k^2$ is negative
Best Answer
Your fraction suddenly disappeared and you forgot the $4$. Your quadratic should be: $$ 4(\tfrac{x}{k}+k) =x^2 \implies \frac{4}{k}x + 4k = x^2 \implies x^2 - \frac{4}{k}x - 4k = 0 $$ Since the line is tangent, the system we are solving should have exactly one solution, so we know that the discriminant is zero. Thus: \begin{align*} 0 = (-\tfrac{4}{k})^2 - 4(1)(-4k) = \frac{16}{k^2} + 16k &\implies 16k = -\frac{16}{k^2} \\ &\implies k = \frac{-1}{k^2} \\ &\implies k^3 = -1 \\ &\implies k = -1 \end{align*}