Widest parabola inscribed in a hyperbola

Calculus Level pending

Consider the hyperbola y 2 − x 2 = 1 y^2 - x^2 = 1 . We want to inscribe a parabola within the upper branch of the hyperbola, such that the parabola is tangent to the hyperbola at the hyperbola vertex ( 0 , 1 ) (0, 1) . If the equation of the parabola is

y = 1 + a x 2 y = 1 + a x^2

where a > 0 a \gt 0 , find the minimum value of a a , so that the parabola is truly inscribed in the hyperbola.


The answer is 0.5.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Tom Engelsman
May 8, 2021

Let the parabola be rewritten as x 2 = y − 1 a \large x^2 = \frac{y-1}{a} and subsititute this into the hyperbola equation:

y 2 − ( y − 1 a ) = 1 ⇒ a y 2 − y + ( 1 − a ) = 0 ⇒ y = 1 ± 1 − 4 ( a ) ( 1 − a ) 2 a = 1 ± 1 − 4 a + 4 a 2 2 a = 1 ± ( 1 − 2 a ) 2 2 a \large y^2 - (\frac{y-1}{a}) = 1 \Rightarrow ay^2 - y + (1-a) = 0 \Rightarrow y = \frac{1 \pm \sqrt{1 - 4(a)(1-a)}}{2a} = \frac{1 \pm \sqrt{1-4a+4a^2}}{2a} = \frac{1 \pm \sqrt{(1-2a)^2}}{2a} .

The least value of a a occurs when the discriminant equals zero ⇒ a = 1 2 . \Rightarrow \boxed{a = \frac{1}{2}}.

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