Angle two curves

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Loki123

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At what angle do curves y^2=4a(a-x) and y^2=4b(b+x) cross each other. a and b are real numbers.
I tried to get coefficients and then calculate the angle but It didn't work. IMG_20220424_124517.jpg
 
If you swap 'x' and 'y' the angle won't change, but the derivatives will be easier to compute.
 
[imath]4a^2 - 4ay = x^2[/imath] and [imath]4b^2+4by = x^2[/imath] looks easier to differentiate. Swapping [imath]x[/imath] and [imath]y[/imath] simply reflects the curves relative to the diagonal [imath]y=x[/imath], but reflection preserves angles. You might want to watch the sign of the angle though.
 
Loki123 said:
Why?
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Maybe not. Nothing in the problem statement says that you need to compute a signed angle. Sometime the angle between two vectors is assigned positive values for counterclockwise direction from the first to the second and negative values for clockwise directions. This does not seem to be the case here, so please disregard my post.
 
blamocur said:
Maybe not. Nothing in the problem statement says that you need to compute a signed angle. Sometime the angle between two vectors is assigned positive values for counterclockwise direction from the first to the second and negative values for clockwise directions. This does not seem to be the case here, so please disregard my post.
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I meant, does switching x and y influence the sign?
 
You'll have to consider 2 cases:
1) Notice that for [imath]a=-b[/imath]. Both curves are exactly the same. This is trivial.
2) For [imath]a\neq -b[/imath], the intersecting points are:
[math]4a^2-4ax=4b^2+4bx\\ (4a^2-4b^2)=(4a+4b)x\\ \frac{(2a-2b)\cancel{(2a+b)}}{2\cancel{(2a+2b)}}=x\\ x=a-b\\ \implies y^2=4a(a-x)=4a(a-(a-b))=4ab\implies y= \pm 2\sqrt{ab}\\ \therefore\text{the intersecting points are } \{a-b,\,\pm 2\sqrt{ab}\}[/math]
 
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BigBeachBanana said:
You'll have to consider 2 cases:
1) Notice that for [imath]a=-b[/imath]. Both curves are exactly the same. This is trivial.
2) For [imath]a\neq -b[/imath], the intersecting points are:
[math]4a^2-4ax=4b^2+4bx\\ (4a^2-4b^2)=(4a+4b)x\\ \frac{(2a-2b)\cancel{(2a+b)}}{2\cancel{(2a+2b)}}=x\\ x=a-b\\ \implies y^2=4a(a-x)=4a(a-(a-b))=4ab\implies y= \pm 2\sqrt{ab}\\ \therefore\text{the intersecting points are } \{a-b,\,\pm 2\sqrt{ab}\}[/math]
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i am not sure i understand where this even came from
 
blamocur said:
If you swap 'x' and 'y' the angle won't change, but the derivatives will be easier to compute.
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dy/dx = tan(t) then dx/dy = cot(t) = tan(pi/2 - t)

If you draw two intersecting curves, the difference of the gradients of the tangents at the point of intersection will not change after reflection.
 
Screen Shot 2022-04-25 at 3.54.44 AM.png
First, you're dividing by 0 here. How do you get 90? Something isn't right.
Second, you have two points of intercepts, therefore you should have 2 angles.
 
Loki123 said:
Isn't tan 90 degrees 1/0?
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Yes, you're right. I was following your work and did not see tan(x) anywhere. I did it with implicit differentiation, the derivative is a bit cleaner IMO.
[math]y_1^2=4a(a-x)\\ 2y\cdot y_1'=-4a\\ y_1'=\frac{-2a}{y}\\[/math]Similarly,
[math]y_2^2=4b(b+x)\\ 2y\cdot y_2'=4b\\ y_2'=\frac{2b}{y}\\[/math]
First point of intercept, [imath](a-b,2\sqrt{ab})[/imath]
[math]y_1'=\frac{-2a}{y}=\frac{-2a}{2\sqrt{ab}}=\frac{-a}{\sqrt{ab}}[/math][math]y_2'=\frac{2b}{y}=\frac{2b}{2\sqrt{ab}}=\frac{b}{\sqrt{ab}}[/math][math]\tan(\theta_1)=\frac{\frac{-a}{\sqrt{ab}}-\frac{b}{\sqrt{ab}}}{1+\frac{-a}{\sqrt{ab}}\cdot \frac{b}{\sqrt{ab}}}\implies \theta_1=90\degree[/math]For the second intercept, plug- in [imath]y=-2\sqrt{ab}[/imath] into the derivatives for the second angle.
 
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