Question

The sum of the distance of a moving points from the points (3, 0) and (-3, 0) is always equal to 12. Find the equation to the locus and identify the conic represented by the equation.

Answer 1

Coordinates Geometry (2D)

Given: the sum of the distance of a moving point from the points (3, 0) and (- 3, 0) is always equal to 12.

To find: the equation of the locus and the conic identified by the equation

Solution: Let the moving point be (p, q)

Its distances from (3, 0) and (- 3, 0) are

$\quad l_{1}=\sqrt{(p-3)^{2}+(q-0)^{2}}$ units

$\Rightarrow l_{1}=\sqrt{p^{2}-6p+9+q^{2}}$ units

$\quad l_{2}=\sqrt{(p+3)^{2}+(q-0)^{2}}$ units

$\Rightarrow l_{2}=\sqrt{p^{2}+6p+9+q^{2}}$ units

By the given condition,

$\quad l_{1}+l_{2}=12$

$\Rightarrow \sqrt{p^{2}-6p+9+q^{2}}+\sqrt{p^{2}+6p+9+q^{2}}=12$

$\Rightarrow \sqrt{p^{2}-6p+9+q^{2}}=12-\sqrt{p^{2}+6p+9+q^{2}}$

$\Rightarrow p^{2}-6p+9+q^{2}=144-24\sqrt{p^{2}+6p+9+q^{2}}+p^{2}+6p+9+q^{2}$

$\Rightarrow 24\sqrt{p^{2}+6p+9+q^{2}}=12(12+p)$

$\Rightarrow 2\sqrt{p^{2}+6p+9+q^{2}}=12+p$

$\Rightarrow 4(p^{2}+6p+9+q^{2})=144+24p+p^{2}$

$\Rightarrow 4p^{2}+24p+36+4q^{2}=144+24p+p^{2}$

$\Rightarrow 3p^{2}+4q^{2}=108$

$\Rightarrow \frac{p^{2}}{36}+\frac{q^{2}}{27}=1$

The required locus is

$\quad \frac{x^{2}}{36}+\frac{y^{2}}{27}=1$

This is a parabola.