- Home
- Standard 11
- Mathematics
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
$\frac{1}{a} - \frac{1}{{a'}} = \frac{1}{b} - \frac{1}{{b'}}$
$\frac{1}{a} + \frac{1}{{a'}} = \frac{1}{b} + \frac{1}{{b'}}$
$\frac{1}{a} + \frac{1}{b} = \frac{1}{{a'}} + \frac{1}{{b'}}$
None of these
Solution
(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 $\frac{1}{b} – \frac{1}{{b'}} = \left( {\frac{1}{a} – \frac{1}{{a'}}} \right)$.
