Find the conditions of $a, b$ and $c$ so the line is tangent to the parabola $y=x^2$

Question

I tried a method for finding a tangent to a parabola described in a book. The example from the book was $y=2x-1$ is tangent to $y=x^2$. I solved it by plugging $y$ in the 2nd equation with $y$ from the 1st to find the roots $x=1$.

I tried the same method to a harder problem and got stuck. Find the conditions on $a, b$ and $c$ so the line (1) $$ax+by+c=0$$ tangents the parabola (2) $$y=x^2$$ I tried differentiating with respect to $x$ but wasn't sure what to do with the derivative. I would like to learn to solve this type of problem using the derivative.

If I plug (1) into (2) I get $$\frac{-ax-c}{b}=x^2$$ How can I continue from here?

Answer 1

The parabola is given as $y = x^2$, therefore

$y' = 2 x = \text{slope of tagent line}$

Therefore, if the tangent line is tangent to the curve $y = x^2$ at $x = x_1 $ then it slope must be

$ m = 2 x_1 $

Further, this tangent line passes through the point $(x_1, x_1^2) $

Therefore, its equation is

$ y - x_1^2 = m (x - x_1 ) $

i.e.

$ -m x + y + m x_1 - x_1^2 = 0 $

but $m = 2 x_1$, therefore, the equation of the tangent line is

$ -2 x_1 x + y + 2 x_1^2 - x_1^2 = 0 $

which simplifies to

$ -2 x_1 x + y + x_1^2 = 0 $

If $k$ is an arbitrary nonzero constant, then

$ a = -2 k x_1 , b = k , c = k x_1^2 $

So that the triple

$(a, b,c) = k ( -2 t , 1, t^2 ) $

for arbitrary $k \in \mathbb{R} , k \ne 0 $ and $t \in \mathbb{R}$.

Answer 2

We may assume that $b\ne 0$ since no line with equation $a x+c=0$ is tangent to the parabola.

Now we just need to solve the system of equations to see what condition on $a$, $b$ and $c$ is necessary to get exactly one solution:

$$y=x^2$$ $$y=\frac{-ax-c}{b}$$

so $$x^2 = \frac{-ax-c}{b}$$ which is $$b x^2+ax+c=0$$

For a quadratic to have exactly on solution we need its discriminant to be $0$, that is $$a^2-4bc=0$$

So the answer is $$a^2=4bc$$ with $b\ne 0$.