Question

Find the locus of a point which moves as sum of its distance from point (1, 0) and (-1,0) remains 3. Which curve is this locus?

Answer 1

Let $P(h,k)$ is any point such that sum of whose distance from $A(1,0)$ and $B(-1,0)$ remains $3$.
According to questions,
$PA+PB=3$
$\Rightarrow \sqrt{(h-1{)}^2+k^2}+\sqrt{(h+1{)}^2+k^2}=3$
$\Rightarrow \sqrt{(h-1{)}^2+k^2}=3-\sqrt{(h+1{)}^2+k^2}$
Squaring on both sides,
$(h-1{)}^2+k^2=9+(h+1{)}^2+k^2-6\sqrt{(h+1{)}^2+k^2}$
$\Rightarrow h^2+1-2h+k^2=9+h^2+1+2h+k^2-6\sqrt{(h+1{)}^2+k^2}$
$\Rightarrow -2h-2h-9=-6\sqrt{(h+1{)}^2+k^2}$
$=\Rightarrow 4h+9=6\sqrt{(h+1{)}^2+k^2}$
squaring on both sides,
$16h^2+81+72h=36\left[\right(h+1{)}^2+k^2]$
$=\Rightarrow 16h62+72h+81-36[h^2+2h+1+k^2]=0$
$=\Rightarrow 16h62+72h+81-36h^2-72h-36-36k^2=0$
$=\Rightarrow -20h^2+36k^2=45$
$=\Rightarrow \frac{20h^2}{45}+\frac{36k^2}{45}=1$
$=\Rightarrow \frac{h^2}{\left(9/4\right)}+\frac{k^2}{\left(5/4\right)}=1$
$\therefore$ Locus of point $(h,k)$ is $\frac{x^2}{\left(9/4\right)}+\frac{y^2}{\left(5/4\right)}=1$, which is required equation of ellipse.