(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider a circle with centre (2a, 0) with radius such that cuts in right angle to the ellipse [tex] b^2 x^2 + a^2 y^2 = a^2 b^2[/tex]. Find the radius

2. Relevant equations

[tex] (x-2a)^2 + y^2 = r^2 [/tex]

[tex] b^2 x^2 + a^2 y^2 = a^2 b^2[/tex]

Answer: [tex] r^2 = \frac{3}{4} (3a^2 + b^2)[/tex]

3. The attempt at a solution

[tex] (x-2a)^2 + y^2 = r^2 [/tex]

[tex] b^2 x^2 + a^2 y^2 = a^2 b^2[/tex]

First obtain the derivates:

[tex] y' = -\frac{b^2 x}{a^2 x} [/tex]

[tex] y' = \frac{2a-x}{y}[/tex]

equal the derivates to -1 and 1 to obtain right angle

[tex] b^2 x = a^2 y [/tex]

[tex] y = 2a-x[/tex]

Now theyre perpendicular. Then solve for x and y, to obtain intersection points:

[tex] y = \frac{2ab}{a^2+b^2} [/tex]

[tex] x = \frac{2a^3}{a^2+b^2} [/tex]

In the following step i think i made a mistake but i dont know which or why. I substituted the x and y values in the circle equation, the answer isnt right.

[tex] r^2 = (\frac{2a^3}{a^2+b^2} - 2a)^2 +(\frac{2ab}{a^2+b^2})^2 [/tex]

[tex] r^2 = \frac{4a^2 b^4 + 4a^2 + b^2}{(a^2+b^2)^2} [/tex]

Could you help me? Thankyou

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# Homework Help: Difficult Derivation Excercise

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