Let $y=ax^2+bx+c$ be a parabola and roots of parabola be $x_1,x_2$. Show that If the tangents drawn from the points where the parabola intersects the $x$-axis are perpendicular to each other, the discriminant of the parabola equation is $1$.
My solution is below:
Product of slope of tangents must be $-1$ . $$-1=(2ax_1+b)(2ax_2+b)=b^2+2ab(x_1+x_2)+4a^2(x_1x_2)=b^2+2ab(-\frac{b}{a})+4a^2(\frac ca)=4ac-b^2=-\Delta.$$
But I use derivative of $y=ax^2+bx+c$. But I am looking for not using derivative.
Observe that solving a line that is tangent with the parabola will have only one solution, which will also be the point of contact.
Let $y = mx + c_t$ be the tangent line, where $m$ is the slope of this line. Solving with the equation of parabola gives us:
$$mx + c_t = ax^2 + bx +c$$
$$\Rightarrow ax^2 + \left(b-m\right)x +c-c_t = 0$$
Since this is a quadratic equation, it has two roots, but since there is only one solution, the root is repeated. It means that the discriminant of the quadratic equation is $0$.
Let the point of contact be $\left(x_0, y_0\right)$.
$$\Rightarrow x_0 = -\frac{b-m}{2a}$$ $$\Rightarrow m = 2ax_0+b$$
Now, you know the slope of the tangent of a parabola to any point without using derivatives.