Difficult Derivation Excercise

[alba_ei]

Homework Statement

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

Homework Equations

$$(x-2a)^2 + y^2 = r^2$$
$$b^2 x^2 + a^2 y^2 = a^2 b^2$$
Answer: $$r^2 = \frac{3}{4} (3a^2 + b^2)$$

The Attempt at a Solution

$$(x-2a)^2 + y^2 = r^2$$
$$b^2 x^2 + a^2 y^2 = a^2 b^2$$

First obtain the derivates:
$$y' = -\frac{b^2 x}{a^2 x}$$
$$y' = \frac{2a-x}{y}$$

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

$$b^2 x = a^2 y$$
$$y = 2a-x$$

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

$$y = \frac{2ab}{a^2+b^2}$$
$$x = \frac{2a^3}{a^2+b^2}$$


In the following step i think i made a mistake but i don't know which or why. I substituted the x and y values in the circle equation, the answer isn't right.

$$r^2 = (\frac{2a^3}{a^2+b^2} - 2a)^2 +(\frac{2ab}{a^2+b^2})^2$$

$$r^2 = \frac{4a^2 b^4 + 4a^2 + b^2}{(a^2+b^2)^2}$$

Could you help me? Thankyou

 

[eigenglue]

Ok first off, if two curves intersect at right angles, then the derivative of one is going to be equal to the negative reciprocal of the other. So if f(x) and g(x) intersect at point a, they interstect at right angles if f'(a)=-1/g'(a). So like if f'(a) is 2, then g'(a) would be -1/2, etc. etc.

Second, of the two relavant equations you listed, on is an ellipse and the other is a circle with center outside the ellipse. Graphically, you know that the circle intersect the ellipse at either zero, one, or two points. In the case of one point, the intersection is tangental, so that's not what you want. Also, graphically, you should be able to see that the case of two point can only exist for values of r such that r > a.

So given that, first solve those two equations for the two points (x,y), each in terms of r, where they intersect. Graphically, you should be able to see both points will have the same value of x and the values of y will be plus and minus something. So solve that, giving you the intersection points x and y in terms of r.

Then determine the derivatives of each of the curves at that point, also in terms of r. Then, in order to get the value r that you are looking for, set the derivative of one equal to the negative reciprocal of the other, and solve for r.

 

[alba_ei]

Okay in this attempt of solution first i obtain the derivate of circle and ellipse.

Circle:

$$(x-2a)^2 + y^2 = r^2$$

Ellipse
$$b^2 x^2 + a^2 y^2 = a^2 b^2$$

Derivate of circle
$$y' = \frac{2a-x}{y}$$

Derivate of ellipse
$$y' = -\frac{b^2 x}{a^2 y}$$

After i equal the derivate of circle to 1 and the derivate of the ellipse to -1 i solve for y and x to find the intersection point.

$$\frac{2a-x}{y} = 1$$

$$-\frac{b^2 x}{a^2 y} = -1$$

$$y = \frac{2ab^2}{a^2+b^2}$$

$$x = \frac{2a^3}{a^2+b^2}$$

so then substituted the value of x and y in the circle equation

$$r^2 = (\frac{2a^3}{a^2+b^2} - 2a)^2 +(\frac{2ab^2}{a^2+b^2})^2$$

$$r^2 = \frac{8a^2 b^4}{(a^2+b^2)^2}$$

But after all that work I am surprised that the answer doesn't match to the one from the book that is

$$r^2 = \frac{3}{4} (3a^2 + b^2)$$

That ansewr even contain fractions and is totally different from mine. I've tried this problem several times but the ansewer still the same, where is my eeror.

 

[huyen_vyvy]

no, when you have the derivatives of the circle and the ellipse, you don't assign 1 and -1 to them. Just multiply them. what you'll get is something like: Ax^2+Bx+Cy^2=0, now solve this quadric equation for x in terms of y, and plug them back into the equations.