Drawing two perpendicular tangent line from the origin to $y=x^2-2x+a$

Solution 1:

You don't need to calculate the tangent lines concretely, you can do it by the process of elimination:

The three values $-\frac{5}{4}, -\frac{3}{4}$ and $\frac{3}{4}$ are not possible. Why? The peak is then below $0$. That means that one tangent has to point below the $x$-axis, however, the other tangent line has to reach the left side of the parabola and therefore go beyond the $y$-axis to the left. That means no angle of $90$ degree is possible.

In fact, for the first two values, you cannot even draw any tangent from $(0,0)$ as this point will lie above the parabola.

That means the only possibility is $a = \frac{5}{4}$. I've added an image of GeoGebra to show that this really works out

Solution 2:

Let the tangent be $y=mx$, (as it passes through the origin), then we have the quadratic $x^2-(m+2)x+4-4a=0.$ In order for the given line be tangent to the given parabola, it must intersect at only one point, and hence ($B^2-4AC=0).$ So we get a quadratic for $m$ as $m^2+2m-4(1-a)=0.$ This means that two tangents are possible, and they will be perpendicular if $m_1m_2=-1 \implies 4(1-a)=-1 \implies a=5/4$ (product of roots of a quadratic $ax^2+bx+c = \frac{c}{a}$)

Solution 3:

Let $f(x)=x^2-2x+a\require{cancel}$.

A line passing through $(0,0)$ is of the form $y=\lambda x$. If such a line is tangent to the graph of $f$ at some point $(\alpha,\beta)$, then:

1. $\beta=f(\alpha)=\alpha^2-2\alpha+a$;
2. $\lambda=f'(\alpha)=2\alpha-2$.

So, you have$$\beta=\lambda\alpha=2\alpha^2-2\alpha\quad\text{and}\quad\beta=\alpha^2-2\alpha+a,$$and therefore $2\alpha^2\cancel{-2\alpha}=\alpha^2\cancel{-2\alpha}+a$. So, $\alpha=\sqrt a$ or $\alpha=-\sqrt a$. In the first case, $\beta=2a-2\sqrt a$, and in the second case $\beta=2a+2\sqrt a$. In the first case, the slope of the tangent line will be $2\sqrt a-2$, and in the second case it will be $-2\sqrt a-2$. The only situation in which the tangent lines will be orthogonal, that is, the only case in which the product of these numbers is $-1$, is when $\sqrt a=\pm\sqrt{\frac54}$. Therefore, $a=\frac54$.