Question

The base $BC$ $(=2a)$ of a triangle $ABC$ is fixed; the axes being $BC$ and a perpendicular to it through its middle point, find the locus of the vertex $A$, when $m$ times the square of one side added to $n$ times the square of the other side is equal to a constant quantity $c^2$.

Answer 1

Let origin be the mis point of $BC$ and the point $A$ be $(h,k)$

In $△ACD{AC}^2={CD}^2+{AD}^2$

In $△ABD{AB}^2={BD}^2+{AD}^2$

${AB}^2={(h+a)}^2+k^2.....\left(i\right)$

Given $m{AC}^2+n{AB}^2=c^2$ ...... $\left(ii\right)$

using $\left(i\right)$ and $\left(ii\right)$

$m({(h-a)}^2+k^2)+n({(h+a)}^2+k^2)=c^{2}m(h^2+a^2-2ah+k^2)+n(h^2+a^2+2ah+k^2)=c^2(m+n)(h^2+a^2+k^2)+2ah(n-m)=c^2$

Replacing $h$ by $x$ and $y$ by $k$

$(m+n)(x^2+a^2+y^2)+2ax(n-m)=c^2$

is the required locus of $P$