Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow y = 2x + a^2 - 2a$
The equation for the normal line is: $y - a^2 = -\frac{1}{2}(x - a) \rightarrow y = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
Now to determine the point of intersection we just see when $f(x)$ and the normal line is equal, i.e.
$x^2 = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
But this seems like a nonsense equation...

The tangent slope of $f(x)$ at $x = a$ is $f'(a) = 2a$, so you will need to amend the tangent and normal line accordingly.
Normal line is $y - a^2 = -\dfrac{1}{2a}(x - a)$ or $y = -\dfrac{x}{2a}+\dfrac{1}{2}+a^2$.
Set $f(x) = y$, so $x^2 = -\dfrac{x}{2a}+\dfrac{1}{2}+a^2$ or $2ax^2 + x - (a + 2a^3) = 0$. Finish with the quadratic formula
$x = \dfrac{-1 \pm \sqrt{1+4(2a)(a+2a^3)}}{4a} = \dfrac{-1 \pm \sqrt{16a^4 + 8a^2 + 1}}{4a}=\dfrac{-1 \pm \sqrt{(4a^2 + 1)^2}}{4a}$
