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I was solving some problems on parabola. I saw a question and solved it, but my solution was way too big. The question was:

If $$\left(\frac{a}{b}\right)^{1/3}+\left(\frac{b}{a}\right)^{1/3} = \frac{\sqrt{3}}{2}$$ Then angle of intersection of parabola $y^2=4ax$ and $x^2=4by$ at a point other than origin is $\ldots$

My solution was too lengthy. I found the point of intersection of parabolas , got tangents, and then found angle as $\dfrac{\pi}{3}$ . Can anybody suggest a shorter method ?

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  • $\begingroup$ But $(a/b)^{\frac13} +(b/a) ^{\frac13} $ cannot be $ <2\, $!! $\endgroup$ – Narasimham Jan 6 '17 at 9:09
  • $\begingroup$ What was $ a/b $ value when you got an included angle $π/3$ @Dinesh? $\endgroup$ – Narasimham Jan 7 '17 at 8:15
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We cannot find two real focal lengths $a,b$ satisfying the given relation. Minimum value of $ (a/b)^3 +(b/a)^3 $ is $=2$ at $ (a/b) = 1 $ (when two identical parabola axes are orthogonally placed).

AnamolousProblFormln

So instead we try to find what may have lead to such an anomalous problem formulation.

Parabola equations $$ y^ 2 = 4 a x,\, x^2= 4 b y\, ;\tag1$$

Points of intersection $$(x,y) = ( 4 a ^{\frac13} b^{\frac23}, 4 a ^{\frac23} b^{\frac13} \,) ; \tag2$$

Slopes at parabolas intersection

$$\left( \frac{2a}{y},\frac {x}{2b} \right) = \left( \frac{(a/b)^{^ \dfrac13}}{2} , 2 (a/b)^ {\dfrac13} \right) = (m_a,m_b); \tag3$$

Tangent of angle $\varphi_{ab}$ between parabola tangents at the intersection point:

$$ \dfrac{m_a-m_b}{1+ m_b m_a} = \dfrac { 3/2 } {{(a/b)}^{\frac13} +{(b/a)}^{\frac13} }; \tag{4}$$

$$ \tan \varphi_{ab} = \dfrac { 3/2 } {{(a/b)}^{\frac13} +{(b/a)}^{\frac13} }; \tag{5}$$

which should be the guiding criterion to set up such a problem.

In a more realistic particular case we take $( a=1,\,b=2) $ The figure verifies computed intersection point and $\varphi_{ab} \approx \tan^{-1} 0.7304 \approx 36.14^0$ graphically a bit less than maximum possible.

 2_Ortho_Parabolas_Cutting

EDIT 1:

Equation(5) above offers some hind/insight. Right hand side is bounded on one side to minimum value $2$ , $\tan $ function has range $ (-\infty,\infty),$ so we cannot have $\varphi_{ab}$ more than $ \tan^{-1}\frac34 \approx 36.87^0 $ when two identical hyperbolae intersect.

Had the problem been set up assuming a solution value $ 0< \varphi_{ab} < 36.87 ^0 $ like e.g.,

$$ (a/b)^{\frac13} +(b/a) ^{\frac13} = 2.5, \tag{6} $$

the problem posing would have been OK.. resulting in two solutions for this case with

$$ a/b = 8, \, 1/8, \, \varphi_{ab} =30.964^0 $$

In Conclusion

Two parabolas whose vertices touch coordinate axes at origin cannot intersect at acute angles in excess of $ \tan^{-1}\frac34 \approx 36.87^ 0= \sin^{-1}\dfrac35. $

EDIT 2:

Maybe we can attempt a generalization:

A single $C_1$ parameter family of curves tangential to x axis at origin cuttng another $C_2$ parameter curves tangent to y-axis at origin forms a maximum angle between tangents when $ C_1=C_2.$

In case of circles they all cut orthogonally.

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  • $\begingroup$ s/hyperbola/parabola/ $\endgroup$ – ccorn Jun 29 '17 at 20:29
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Use the optical properties of the parabola. Let $\Gamma$ be a parabola with vertex in $V$ and let $P\neq V$ a point on $\Gamma$, $l$ the tangent to $\Gamma$ in $P$. Given that $P_\perp$ is the projection of $P$ on the axis $a$ of the parabola, we have that $a\cap l$ is just the symmetric of $P_\perp$ with respect to $V$.

With a few calculations, this just gives that your first formula is the sine of the angle between the tangents.

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