Bizarre Hyperbola

Calculus Level 4

The hyperbola above has equation 2 x 2 + 10 x y + 8 y 2 + 4 x + 2 y + 1 = 0 2x^2 + 10xy + 8y^2 + 4x + 2y + 1 = 0 .

Let m 1 m_{1} and m 2 m_{2} be the slopes of the asymptotes (shown in green) such that ∣ m 1 ∣ > ∣ m 2 ∣ |m_{1}| > |m_{2}| , and let m 3 m_{3} be the slope of the line that passes through both points, where the tangents have infinite slopes.

Find the value of 1 ∣ m 1 ∣ − ∣ m 2 ∣ − ∣ m 3 ∣ \dfrac{1}{|m_{1}| - |m_{2}| - |m_{3}|} .


The answer is 8.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

By implicit differentiation of 2 x 2 + 10 x y + 8 y 2 + 4 x + 2 y + 1 = 0 2x^2 + 10xy + 8y^2 + 4x + 2y + 1 = 0 , we can calculate f ′ ( x ) f'(x) as:

4 x + 10 ( y + x ⋅ f ′ ( x ) ) + 16 y ⋅ ( f ′ ( x ) ) + 4 + 2 ( f ′ ( x ) ) = 0 4x + 10(y + x\cdot f'(x)) + 16y\cdot (f'(x)) + 4 + 2(f'(x)) = 0

f ′ ( x ) ⋅ ( 16 y + 10 x + 2 ) = − 10 y − 4 x − 4 f'(x) \cdot (16y +10x + 2) = -10y - 4x - 4

f ′ ( x ) = − ( 5 y + 2 x + 2 ) 8 y + 5 x + 1 f'(x) = \dfrac{-(5y + 2x + 2)}{8y + 5x +1}

Upon the vertices of the hyperbola, the tangents touching them will have the slope of infinity as the x-values approach local minimum and maximum. Therefore, for infinite slope, the denominator of f ′ ( x ) f'(x) reaches 0 : 8 y + 5 x + 1 = 0 8y + 5x + 1 = 0 is the line passing through these vertices.

Then y = − 5 x 8 − 1 8 y = -\dfrac{5x}{8} - \dfrac{1}{8}

Thus, m 3 = − 5 8 m_{3} = -\dfrac{5}{8} .

Now for the two asymptotes, we know that the slope of these lines will touch the hyperbola as the x-values reach plus or minus infinity, so we can apply the limit to f ′ ( x ) f'(x) :

lim ⁡ x → ∞ f ′ ( x ) = lim ⁡ x → ∞ − ( 5 y + 2 x + 2 ) 8 y + 5 x + 1 \lim_{x\to\infty} f'(x) = \lim_{x\to\infty} \dfrac{-(5y + 2x + 2)}{8y + 5x +1}

And ( − 5 8 ) ( 8 y + 5 x + 1 ) = − 5 y − 25 x 8 − 5 8 (\dfrac{-5}{8})(8y + 5x +1) = -5y - \dfrac{25x}{8} - \dfrac{5}{8}

Hence, by polynomial division,

lim ⁡ x → ∞ f ′ ( x ) = lim ⁡ x → ∞ [ − 5 8 + 25 x 8 + 5 8 − 2 x − 2 8 y + 5 x + 1 ] = lim ⁡ x → ∞ [ − 5 8 + 9 x 8 + − 11 8 8 y + 5 x + 1 ] \lim_{x\to\infty} f'(x) = \lim_{x\to\infty} [\dfrac{-5}{8} + \dfrac{ \dfrac{25x}{8} + \dfrac{5}{8}- 2x - 2}{8y + 5x +1}] = \lim_{x\to\infty} [\dfrac{-5}{8} + \dfrac{ \dfrac{9x}{8} + \dfrac{-11}{8} }{8y + 5x +1}]

Then, from the equation, we can solve for y-value as the following:

8 y 2 + ( 10 x + 2 ) y + ( 2 x 2 + 4 x + 1 ) = 0 8y^2 + (10x + 2)y + (2x^2 + 4x +1) = 0

y = − ( 10 x + 2 ) ± ( 10 x + 2 ) 2 − 4 ⋅ 8 ⋅ ( 2 x 2 + 4 x + 1 ) 2 ⋅ 8 = − ( 5 x + 1 ) ± 9 x 2 − 22 x − 7 8 y = \dfrac{-(10x + 2) \pm \sqrt{(10x + 2)^2 - 4\cdot 8\cdot (2x^2 + 4x +1) }}{2\cdot 8} = \dfrac{-(5x + 1) \pm \sqrt{9x^2 - 22x -7}}{8}

Thus, 8 y + 5 x + 1 = ± 9 x 2 − 22 x − 7 8y + 5x +1 = \pm \sqrt{9x^2 - 22x -7} .

As a result, lim ⁡ x → ∞ [ − 5 8 + 9 x 8 + − 11 8 8 y + 5 x + 1 ] = lim ⁡ x → ∞ [ − 5 8 ± 9 x 8 + − 11 8 9 x 2 − 22 x − 7 ] \lim_{x\to\infty} [\dfrac{-5}{8} + \dfrac{ \dfrac{9x}{8} + \dfrac{-11}{8} }{8y + 5x +1}] = \lim_{x\to\infty} [\dfrac{-5}{8} \pm \dfrac{ \dfrac{9x}{8} + \dfrac{-11}{8} }{\sqrt{9x^2 - 22x -7}}]

The square root result of 9 x 2 − 22 x − 7 \sqrt{9x^2 - 22x -7} will have a highest degree of 3 x 3x , and as x x approaches infinity, the values of other lesser degrees will be so infinitesimal and, thereby, negligible.

Thus, lim ⁡ x → ∞ [ − 5 8 ± 9 x 8 + − 11 8 9 x 2 − 22 x − 7 ] = lim ⁡ x → ∞ [ − 5 8 ± 9 x 8 + − 11 8 3 x ] = − 5 8 ± 3 8 = − 1 4 or − 1 \lim_{x\to\infty} [\dfrac{-5}{8} \pm \dfrac{ \dfrac{9x}{8} + \dfrac{-11}{8} }{\sqrt{9x^2 - 22x -7}}] = \lim_{x\to\infty} [\dfrac{-5}{8} \pm \dfrac{ \dfrac{9x}{8} + \dfrac{-11}{8} }{3x}] = \dfrac{-5}{8} \pm \dfrac{3}{8} = \dfrac{-1}{4} \text{or} -1

Hence, ∣ m 1 ∣ = 1 |m_{1}| = 1 ; ∣ m 2 ∣ = 1 4 |m_{2}| = \dfrac{1}{4} .

Finally, 1 ∣ m 1 ∣ − ∣ m 2 ∣ − ∣ m 3 ∣ = 1 1 − 1 4 − 5 8 = 8 \dfrac{1}{|m_{1}| - |m_{2}| - |m_{3}|} = \dfrac{1}{1 - \dfrac{1}{4} - \dfrac{5}{8}} = \boxed{8} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Let V 1 V_1 and V 2 V_2 be the vertices of the hyperbola, let t 1 t_1 and t 2 t_2 be the tangents to the hyperbola at V 1 V_1 and V 2 V_2 , and let ℓ \ell be the line passing through V 1 V_1 and V 2 V_2 . You assume that t 1 t_1 and t 2 t_2 are vertical (by taking the line with slope of infinity), but they are not.

Because of the x y xy term, the hyperbola is slightly rotated, so ℓ \ell is not perfectly horizontal. But ℓ \ell and t 1 t_1 are perpendicular, so t 1 t_1 is not perfectly vertical (and the same for t 2 t_2 ).

We can calculate m 3 m_3 as follows. We know that m 1 = − 1 m_1 = -1 and m 2 = − 1 4 m_2 = -\frac{1}{4} . Let α \alpha be the angle between the line with slope m 1 m_1 and the x x -axis, so tan ⁡ α = 1 \tan \alpha = 1 . Let β \beta be the angle between the line with slope m 2 m_2 and the x x -axis, so tan ⁡ β = 1 4 \tan \beta = \frac{1}{4} .

Let θ \theta be the angle between ℓ \ell and the x x -axis. Note that ℓ \ell bisects the asymptotes, so θ = α + β 2 \theta = \frac{\alpha + \beta}{2} and m 3 = − tan ⁡ θ m_3 = -\tan \theta .

Note that tan ⁡ 2 θ = tan ⁡ ( α + β ) = tan ⁡ α + tan ⁡ β 1 − tan ⁡ α tan ⁡ β = 5 3 . \tan 2 \theta = \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{5}{3}. But tan ⁡ 2 θ = 2 tan ⁡ θ 1 − tan ⁡ 2 θ , \tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}, so 2 tan ⁡ θ 1 − tan ⁡ 2 θ = 5 3 . \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{5}{3}. Solving for tan ⁡ θ \tan \theta , we get tan ⁡ θ = − 3 ± 34 5 . \tan \theta = \frac{-3 \pm \sqrt{34}}{5}.

Since m 3 = − tan ⁡ θ m_3 = -\tan \theta is negative, m 3 = 3 − 34 5 . m_3 = \frac{3 - \sqrt{34}}{5}.

Jon Haussmann - 4 years, 10 months ago

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