Ellipse with two perpendicular tangent: x^2/a^2 + y^2/b^2 = 1

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otavukcuoglu

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The tangents at the points\(\displaystyle \, (a\, \cos(\alpha),\, b\, \sin(\alpha))\, \) and \(\displaystyle \, (a\, \cos(\beta),\, b\, \sin(\beta))\, \) to the ellipse:

. . . . .\(\displaystyle \dfrac{x^2}{a^2}\, +\, \dfrac{y^2}{b^2}\, =\, 1\)

are perpendicular. Show that:

. . . . .\(\displaystyle \tan(\alpha)\, \tan(\beta)\, =\, -\,\dfrac{b^2}{a^2}\)


can you help me to solve this it is very confusing.
 

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otavukcuoglu said:
The tangents at the points\(\displaystyle \, (a\, \cos(\alpha),\, b\, \sin(\alpha))\, \) and \(\displaystyle \, (a\, \cos(\beta),\, b\, \sin(\beta))\, \) to the ellipse:

. . . . .\(\displaystyle \dfrac{x^2}{a^2}\, +\, \dfrac{y^2}{b^2}\, =\, 1\)

are perpendicular. Show that:

. . . . .\(\displaystyle \tan(\alpha)\, \tan(\beta)\, =\, -\,\dfrac{b^2}{a^2}\)


can you help me to solve this it is very confusing.
Click to expand...
Please reply showing your thoughts and efforts so far, so we can see where things have gotten confusing. Thank you! ;)
 
otavukcuoglu said:
The tangents at the points\(\displaystyle \, (a\, \cos(\alpha),\, b\, \sin(\alpha))\, \) and \(\displaystyle \, (a\, \cos(\beta),\, b\, \sin(\beta))\, \) to the ellipse:

. . . . .\(\displaystyle \dfrac{x^2}{a^2}\, +\, \dfrac{y^2}{b^2}\, =\, 1\)

are perpendicular. Show that:

. . . . .\(\displaystyle \tan(\alpha)\, \tan(\beta)\, =\, -\,\dfrac{b^2}{a^2}\)


can you help me to solve this it is very confusing.
Click to expand...
What are the "slopes"of the tangents at those points?

How is the perpendicularity related to the slopes?
 
stapel said:
Please reply showing your thoughts and efforts so far, so we can see where things have gotten confusing. Thank you! ;)
Click to expand...

I know that the product of gradient should be -1 and the coefficient of x gives us the gradients but I can't solve the rest.
 
stapel said:
Please reply showing your thoughts and efforts so far, so we can see where things have gotten confusing. Thank you! :wink:
Click to expand...
I also forgot to mention this the tangent formula of ellipse is that xcos(T)/a+ysin(T)/b=1 when i put these two points in to the formula i can't solve the rest
 
The tangents at the points[FONT=MathJax_Main]([FONT=MathJax_Math-italic]a[FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]b[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT](acos⁡(α),bsin⁡(α)) and [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]β[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]b[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]β[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT](acos⁡(β),bsin⁡(β)) to the ellipse:[/FONT][/FONT]

. . . . .[FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]b[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT]x2a2+y2b2=1

are perpendicular. Show that:

. . . . .[FONT=MathJax_Main]tan[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]tan[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]β[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]b[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main]2[/FONT]tan⁡(α)tan⁡(β)=−b2a2
Click to expand...
otavukcuoglu said:
I also forgot to mention this the tangent formula of ellipse is that xcos(T)/a+ysin(T)/b=1 when i put these two points in to the formula i can't solve the rest
Click to expand...

What is 'T? How is it related to α or ß?
 
otavukcuoglu said:
The tangents at the points\(\displaystyle \, (a\, \cos(\alpha),\, b\, \sin(\alpha))\, \) and \(\displaystyle \, (a\, \cos(\beta),\, b\, \sin(\beta))\, \) to the ellipse: .\(\displaystyle \dfrac{x^2}{a^2}\, +\, \dfrac{y^2}{b^2}\, =\, 1\)
are perpendicular. Show that:.\(\displaystyle \tan(\alpha)\, \tan(\beta)\, =\, -\,\dfrac{b^2}{a^2}\)
can you help me to solve this it is very confusing.
Click to expand...

The parametric equation of the ellipse is:\(\displaystyle r(t) = \left\{ \begin{array}{l}x(t)=a\cos (t)\\y(t)=b\sin (t)\end{array} \right.\) To find slope \(\displaystyle r'(t) = \left\{ \begin{array}{l}x'(t)=-a\sin (t)\\y'(t)=b\cos (t)\end{array} \right.\)

That is \(\displaystyle \dfrac{dx}{dt}=x'(t)=-a\sin(t)~\&~\dfrac{dy}{dt}=y'(t)=b\cos(t)\) so what is slope \(\displaystyle m=\dfrac{dy}{dx}=~?\)
 
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pka said:
The parametric equation of the ellipse is:\(\displaystyle r(t) = \left\{ \begin{array}{l}x(t)=a\cos (t)\\y(t)=b\sin (t)\end{array} \right.\) To find slope \(\displaystyle r'(t) = \left\{ \begin{array}{l}x'(t)=-a\sin (t)\\y'(t)=b\cos (t)\end{array} \right.\)

That is \(\displaystyle \dfrac{dx}{dt}=x'(t)=-a\sin(t)~\&~\dfrac{dy}{dt}=y'(t)=b\cos(t)\) so what is slope \(\displaystyle m=\dfrac{dy}{dx}=~?\)
Click to expand...

This was the whole question it did not mention about slope
 
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