Question

$P$ and $Q$ are two variable points on the axes of $x$ and $y$ respectively such that $|OP|+|OQ|=a$, then the locus of foot of perpendicular from origin on $PQ$ is

• A. $(x-y)(x^2+y^2)=axy$
• B. $(x+y)(x^2+y^2)=axy$
• C. $(x+y)(x^2+y^2)=a(x-y)$
• D. $(x+y)(x^2-y^2)=axy$

Answer 1

The correct option is B $(x+y)(x^2+y^2)=axy$

$\left|P\right|+\left|q\right|=a$
$p^2+q^2+2\left|P\right|\left|q\right|=a^2$
$(p^2+q^2-a^2{)}^2=4{\left|P\right|}^2{\left|q\right|}^2$
$p^4+q^4+a^4+2p^2{q}^2-2a^2{p}^2-2a^2{q}^2=4p^2{q}^2$
$P^4+q^4+a^4+-2p^2{q}^2-2a^2{p}^2-2a^2{q}^2=0$
${(p^2-q^2)}^2+a^2(a^2-2p^2-2q^2)=0\left(1\right)$
y = mx
$m\times \left|\frac{q-0}{0-p}\right|=-1$
$m=\frac{p}{q}-\left(2\right)$
$k=\frac{p}{q}h\left(3\right)$
$\left|\begin{array}{ccc}o& q& 1\\ h& k& 1\\ p& 0& 1\end{array}\right|=0\left(4\right)$
from (4) & (3)
$P=\frac{h-k^2}{h},q=\frac{h-k^2}{k}$
after putting the value of P and q in the $e{q}^n$ (1) and simplilying it
we get,
$(h+k)(h^2+k^2)=qhk$
$(x+y)(x^2+y^2)=axy$