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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.

Dave L. Renfro
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matholympicman
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  • I not sure if it is possible to find the slop of the tangent of a parabola without using derivative (explicitly or implicitly). Maybe it is possible to prove this result using the reflection properties of the parabola, but, again, I could not find any proof of those those properties that are not based in derivatives. – Renato Fernandes Nov 14 '23 at 13:34

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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.

Shubhav Jain
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