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Angle of intersection of curves-

Posted May 5th, 2010 by alpha
in

Show that the curves
$ \frac{x^2}{a^2+\lambda_1}+\frac{y^2}{b^2+\lambda_1}=1 $&
$ \frac{x^2}{a^2+\lambda_2}+\frac{y^2}{b^2+\lambda_2}=1 $ intersect at right angles.

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Comments

I know the concept to be used.

On May 5th, 2010 alpha says:

The angle of intersection of curves refers to the angle between the tangents drawn to the curves at the point of intersection of the curves.

These curves are needed to be solved to get the point of intersection.
On writing the eqn. of the curve in other way ,I got
$ x^2(b^2+\lambda_1) +y^2(a^2+\lambda_1)-(a^2+\lambda_1)(b^2+\lambda_1)=0 $
Similarly , the second eqn. can be written as

$ x^2(b^2+\lambda_2) +y^2(a^2+\lambda_2)-(a^2+\lambda_2)(b^2+\lambda_2)=0 $
How will I solve these( with any trick)?These eqns look too weird and solving them will be very clumsy.
-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On May 8th, 2010 Structure says:

You can answer your question without known the explicit value of common points $ (x_0,y_0) $
Let f and g be implicit local functions such as

$$\frac{x^2}{a^2+\lambda_{1}}+\frac{f^2(x)}{b^2+\lambda_{1}}=1$$
$$\frac{x^2}{a^2+\lambda_{2}}+\frac{g^2(x)}{b^2+\lambda_{2}}=1$$

such as

$$f(x_0)=g(x_0)=y_0$$

The slopes in common point are given by $ f'(x_0) $ and $ g'(x_0) $.
Orthogonality means

$$f'(x_0)g'(x_0)+1=0$$

You have

$$\frac{x_0}{a^2+\lambda_{1}}+\frac{f(x_0)f'(x_0)}{b^2+\lambda_{1}}=0$$
$$\frac{x_0}{a^2+\lambda_{2}}+\frac{g(x_0)g'(x_0)}{b^2+\lambda_{2}}=0$$

or

$$\frac{x_0}{a^2+\lambda_{1}}+\frac{y_0f'(x_0)}{b^2+\lambda_{1}}=0$$
$$\frac{x_0}{a^2+\lambda_{2}}+\frac{y_0g'(x_0)}{b^2+\lambda_{2}}=0$$

but from

$$\frac{x^2_0}{a^2+\lambda_{1}}+\frac{y_0^2}{b^2+\lambda_{1}}=1$$
$$\frac{x^2_0}{a^2+\lambda_{2}}+\frac{y^2_0}{b^2+\lambda_{2}}=1$$

you get by substraction

$$(\lambda_2-\lambda_1)(\frac{x_0^2}{(a^2+\lambda_{1})(a^2+\lambda_{2})}+\frac{y_0^2}{(b^2+\lambda_{1})(b^2+\lambda_{2})})=0$$

equivalent to $ f'(x_0)g'(x_0)+1=0 $

answer

On May 8th, 2010 Structure says:

You can answer your question without known the explicit value of common points $ (x_0,y_0) $
Let f and g be implicit local functions such as

$$\frac{x^2}{a^2+\lambda_{1}}+\frac{f^2(x)}{b^2+\lambda_{1}}=1$$
$$\frac{x^2}{a^2+\lambda_{2}}+\frac{g^2(x)}{b^2+\lambda_{2}}=1$$

such as

$$f(x_0)=g(x_0)=y_0$$

The slopes in common point are given by $ f'(x_0) $ and $ g'(x_0) $.
Orthogonality means

$$f'(x_0)g'(x_0)+1=0$$

You have

$$\frac{x_0}{a^2+\lambda_{1}}+\frac{f(x_0)f'(x_0)}{b^2+\lambda_{1}}=0$$
$$\frac{x_0}{a^2+\lambda_{2}}+\frac{g(x_0)g'(x_0)}{b^2+\lambda_{2}}=0$$

or

$$\frac{x_0}{a^2+\lambda_{1}}+\frac{y_0f'(x_0)}{b^2+\lambda_{1}}=0$$
$$\frac{x_0}{a^2+\lambda_{2}}+\frac{y_0g'(x_0)}{b^2+\lambda_{2}}=0$$

but from

$$\frac{x^2_0}{a^2+\lambda_{1}}+\frac{y_0^2}{b^2+\lambda_{1}}=1$$
$$\frac{x^2_0}{a^2+\lambda_{2}}+\frac{y^2_0}{b^2+\lambda_{2}}=1$$

you get by substraction

$$(\lambda_2-\lambda_1)(\frac{x_0^2}{(a^2+\lambda_{1})(a^2+\lambda_{2})}+\frac{y_0^2}{(b^2+\lambda_{1})(b^2+\lambda_{2})}=0$$

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