If the lines represented by the equation x^2+y^2=c^2((bx+ay)/(ab))^2 form a right angle then prove that:1/a^2+1/b^2+1/c^2=3/c^2?

Jan 1, 2018

Explanation:

The equation ${x}^{2} + {y}^{2} = {c}^{2} {\left(\frac{b x + a y}{a b}\right)}^{2}$ can be expanded as

${a}^{2} {b}^{2} {x}^{2} + {a}^{2} {b}^{2} {y}^{2} = {c}^{2} {b}^{2} {x}^{2} + {c}^{2} {a}^{2} {y}^{2} + 2 a b {c}^{2} x y$

or ${x}^{2} \left({a}^{2} {b}^{2} - {b}^{2} {c}^{2}\right) - 2 a b {c}^{2} x y + {y}^{2} \left({a}^{2} {b}^{2} - {c}^{2} {a}^{2}\right) = 0$

or ${b}^{2} {x}^{2} \left({a}^{2} - {c}^{2}\right) - 2 a b {c}^{2} x y + {a}^{2} {y}^{2} \left({b}^{2} - {c}^{2}\right) = 0$

or $\frac{{b}^{2} \left({a}^{2} - {c}^{2}\right)}{{a}^{2} \left({b}^{2} - {c}^{2}\right)} {x}^{2} - \frac{2 a b {c}^{2}}{{a}^{2} \left({b}^{2} - {c}^{2}\right)} x y + {y}^{2} = 0$

As there is no term containing $x$, $y$ or constant term,

the two lines would be passing trough origin and lines could be $y = {m}_{1} x$ and $y = {m}_{2} x$, and hence the quadratic equation would be

$\left({m}_{1} x - y\right) \left({m}_{2} x - y\right) = 0$ i.e. ${m}_{1} {m}_{2} {x}^{2} - \left({m}_{1} + {m}_{2}\right) x y + {y}^{2} = 0$

Hence comparing the two equations

${m}_{1} {m}_{2} = \frac{{b}^{2} \left({a}^{2} - {c}^{2}\right)}{{a}^{2} \left({b}^{2} - {c}^{2}\right)}$

but as they form right angle ${m}_{1} {m}_{2} = - 1$

i.e. $\frac{{b}^{2} \left({a}^{2} - {c}^{2}\right)}{{a}^{2} \left({b}^{2} - {c}^{2}\right)} = - 1$

or ${b}^{2} {a}^{2} - {b}^{2} {c}^{2} = - {a}^{2} {b}^{2} + {a}^{2} {c}^{2}$

or ${b}^{2} {a}^{2} - {b}^{2} {c}^{2} + {a}^{2} {b}^{2} - {a}^{2} {c}^{2} = 0$

or ${b}^{2} {c}^{2} + {a}^{2} {c}^{2} + {a}^{2} {b}^{2} = 3 {a}^{2} {b}^{2}$

and dividing by ${a}^{2} {b}^{2} {c}^{2}$ we get

$\frac{1}{a} ^ 2 + \frac{1}{b} ^ 2 + \frac{1}{c} ^ 2 = \frac{3}{c} ^ 2$