The condition of the curves a{x^2} + b{y^2} = | vedclass

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Question

(a) Solving for ${x^2},\;{y^2}$;$\left( {\sqrt {\frac{{b' - b}}{{ab' - ba'}}} {\rm{,}}\,{\rm{ }}\sqrt {\frac{{a' - a}}{{a'b - b

Vedclass · Accepted Answer

$\frac{1}{a} - \frac{1}{{a'}} = \frac{1}{b} - \frac{1}{{b'}}$

Vedclass · Answer

(a) Solving for ${x^2},\;{y^2}$;$\left( {\sqrt {\frac{{b' - b}}{{ab' - ba'}}} {\rm{,}}\,{\rm{ }}\sqrt {\frac{{a' - a}}{{a'b - b'a}}} } \right)$ is the intersecting point.

Differentiating $a{x^2} + b{y^2} = 1$, $2ax + 2by\frac{{dy}}{{dx}} = 0$

$ \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_1} = - \frac{{ax}}{{ay}}$

and ${\left( {\frac{{dy}}{{dx}}} \right)_2} = - \frac{{a'x}}{{b'y}}$

and ${\left( {\frac{{dy}}{{dx}}} \right)_1}{\left( {\frac{{dy}}{{dx}}} \right)_2} = - 1$

$ \Rightarrow \frac{{aa'}}{{bb'}}\left( {\frac{{{x^2}}}{{{y^2}}}} \right) = - 1$

or $\frac{{aa'}}{{bb'}}\left( {\frac{{b' - b}}{{a' - a}}} \right) = 1$.

Hence $\frac{1}{b} - \frac{1}{{b'}} = \left( {\frac{1}{a} - \frac{1}{{a'}}} \right)$.

The condition of the curves $a{x^2} + b{y^2} = 1$and $a'{x^2} + b'{y^2} = 1$ to intersect each other orthogonally, is

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