Tangency and Pythagorean Triplets.

#ISI#CoordinateGeometry#PythagoreanTriples#Parabola#Tangent

Problem Statement

Let $a$ and $b$ be rational numbers. Consider the parabola

$$ y=ax^{2}+b $$

and the line

$$ y=b^{2}x-a. $$

Which of the following statements are true?

(A) There exist infinitely many rational pairs $(a,b)$ for which the line is tangent to the parabola.

(B) There exists at least one pair of positive rational numbers $(a,b)$ for which the line is tangent to the parabola.

$$ b^{4}=4a(a+b). $$

(D) There exists exactly one rational pair $(a,b)$ for which the line is tangent to the parabola.

Answer:(A,B,C)

Solution: The line

$$ y=b^{2}x-a $$

is a tangent to the parabola

$$ y=ax^{2}+b $$

if and only if the equation

$$ ax^{2}-b^{2}x+(a+b)=0 $$

has a repeated root. Hence,

$$ \Delta=b^{4}-4a(a+b)=0, $$

or equivalently,

$$ 4a^{2}+4ab-b^{4}=0. $$

Observe that this is a quadratic equation in $a$. Therefore, $a$ is rational if and only if its discriminant is the square of a rational number

$$ 16b^{2}(1+b^{2}) $$

is the square of a rational number.

Now, let

$$ b=\frac{p}{q}, \ \text{where} \ p,q \in \mathbb{Z} \ \text{and gcd}(p,q)=1 $$

and $(p,q,\ell)$ is a Pythagorean triple, that is,

$$ p^{2}+q^{2}=\ell^{2}. $$

Then

$$ 16b^{2}(1+b^{2}) = 16\cdot \frac{p^{2}}{q^{2}} \cdot \frac{p^{2}+q^{2}}{q^{2}} = 16\cdot \frac{p^{2}\ell^{2}}{q^{4}} = \left(\frac{4p\ell}{q^{2}}\right)^{2}, $$

which is indeed the square of a rational number. Consequently, $a$ is rational.

Since there exist infinitely many primitive Pythagorean triples, therefore there are infinitely many rational values of $b$, and hence infinitely many rational pairs $(a,b)$ satisfying the given condition.

Bibliography & References

[1]
ISI,2026