Q.The equation of a hyperbola is $\frac{x^{2}}{36} - \frac{y^{2}}{64} = 1$, then,

Set $3\cdot PA = 2\cdot PB$, square both sides, and simplify using the distance formula.

Let $P=(x,y)$. Given $\dfrac{PA}{PB}=\dfrac23$ where $A=(5,-4)$, $B=(7,6)$, i.e. $3\,PA=2\,PB$, so $9\,PA^2=4\,PB^2$.

$PA^2=(x-5)^2+(y+4)^2 = x^2+y^2-10x+8y+41$

$PB^2=(x-7)^2+(y-6)^2 = x^2+y^2-14x-12y+85$

…

Write $PA:PB=2:3$ as $9\,PA^2=4\,PB^2$ using the distance formula, then simplify.

Let $P=(x,y)$, $A=(2,3)$, $B=(2,-3)$. Given $\dfrac{PA}{PB}=\dfrac{2}{3}$, so $3\,PA = 2\,PB$, i.e. $9\,PA^2 = 4\,PB^2$.

$PA^2 = (x-2)^2+(y-3)^2, \qquad PB^2 = (x-2)^2+(y+3)^2$

So:

$9\big[(x-2)^2+(y-3)^2\big] = 4\big[(x-2)^2+(y+3)^2\big]$

Group the $(x-2)^2$ terms and expand the $y$ terms:

$5(x-2)^2 + \big[9(y-3)^2 - 4(y+3)^2\big] = 0$

$9(y-3)^2 = 9y^2-54y+81$, and $4(y+3)^2=4y^2+24y+36$, so: …

Use the area-of-triangle formula with $P(x,y)$, $A(5,3)$, $B(3,-2)$, set it equal to $9$, and solve.

Let $P=(x,y)$, $A=(5,3)$, $B=(3,-2)$. The area of triangle $PAB$ is

$\text{Area} = \frac{1}{2}\big| x(3-(-2)) + 5(-2-y) + 3(y-3) \big|$

$= \frac{1}{2}\big| 5x - 10 - 5y + 3y - 9 \big| = \frac{1}{2}\big| 5x - 2y - 19 \big|$

Setting the area equal to $9$:

$\frac{1}{2}|5x-2y-19| = 9 \implies |5x-2y-19| = 18$

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Expanding $(PA)^2+(PB)^2=2k^2$ for $A(3,4,5)$, $B(-1,3,-7)$ gives $2x^2+2y^2+2z^2-4x-14y+4z+109-2k^2=0$.

Let $P=(x,y,z)$. Then:

$PA^2 = (x-3)^2+(y-4)^2+(z-5)^2$

$PB^2 = (x+1)^2+(y-3)^2+(z+7)^2$

Expanding each pair of terms:

$(x-3)^2+(x+1)^2 = 2x^2-4x+10$

$(y-4)^2+(y-3)^2 = 2y^2-14y+25$

$(z-5)^2+(z+7)^2 = 2z^2+4z+74$

So:

$PA^2+PB^2 = 2x^2+2y^2+2z^2-4x-14y+4z+(10+25+74)$

$= 2x^2+2y^2+2z^2-4x-14y+4z+109$

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The locus is the ellipse $16(x-2)^2+7y^2=112$ (equivalently $16x^2+7y^2-64x-48=0$).

Let $P=(x,y)$. Then

$PA+PB=\sqrt{(x-2)^2+(y-3)^2}+\sqrt{(x-2)^2+(y+3)^2}=8.$

Move one radical over and square:

$(x-2)^2+(y-3)^2 = 64-16\sqrt{(x-2)^2+(y+3)^2}+(x-2)^2+(y+3)^2.$

The $(x-2)^2$ cancel; $(y-3)^2-(y+3)^2=-12y$, so

$-12y=64-16\sqrt{(x-2)^2+(y+3)^2}\implies 4\sqrt{(x-2)^2+(y+3)^2}=16+3y.$

Square again:

$16\big[(x-2)^2+(y+3)^2\big]=256+96y+9y^2.$ …