Vectors and conic sections - Part 1

Position coordinates of $P$ are $(0,1)$ . Let the position coordinates of $A$ and $B$ be $(h, 1-h)$ and $(k, 1-k)$ respectively.

Then $|\overline {PA}|=\dfrac{5}{12}|\overline {PB}|\implies k=\dfrac{12h}{5}\implies h+k=\dfrac{17h}{5}, hk=\dfrac{12h^2}{5}$ .

Solving the equations of the hyperbola and the straight line we get $(a^2-1)x^2-2a^2x+2a^2=0$ , whose roots are $h, k$ . Therefore $h+k=\dfrac{2a^2}{a^2-1}=\dfrac{17h}{5}, hk=\dfrac{2a^2}{a^2-1}=\dfrac{12h^2}{5}$ . So, $120a^2=289a^2-289\implies 169a^2=289\implies a=\dfrac{17}{13}$ . Therefore $p=17, q=13$ and $p+q=\boxed {30}$ .

The two points $A(x_a, y_a)$ and $B(x_b, y_b)$ satisfy both $\dfrac {x^2}{a^2} - y^2 = 1$ and $x + y = 1 \implies y = 1 -x$ and therefore,

$\begin{aligned} y^2 & = \frac {x^2}{a^2} - 1 \\ a^2(1-x)^2 & = x^2 - a^2 \\ (a^2-1)x^2 - 2a^2x + 2a^2 & = 0 \end{aligned}$

$\implies x = \dfrac {a^2 \pm a\sqrt{2-a^2}}{a^2-1} \implies x_a = \dfrac {a^2 - a\sqrt{2-a^2}}{a^2-1}$ and $x_b = \dfrac {a^2 + a\sqrt{2-a^2}}{a^2-1}$ We note that $\dfrac {PA}{PB} = \dfrac {x_a}{x_b}$ , therefore,

$\begin{aligned} \frac {a^2-a\sqrt{2-a^2}}{a^2+a\sqrt{2-a^2}} & = \frac 5{12} \\ \frac {a-\sqrt{2-a^2}}{a+\sqrt{2-a^2}} & = \frac 5{12} \\ 12a - 12\sqrt{2-a^2} & = 5a + 5\sqrt{2-a^2} \\ 7a & = 17\sqrt{2-a^2} \\ 49a^2 & = 289(2-a^2) \\ a^2 & = \frac {289}{169} \\ \implies a & = \frac {17}{13} \end{aligned}$

Therefore, $p+q = 17+13 = \boxed {20}$ .