Given that line $y=mx+r $ meets the curve $\frac{y^2}{p^2}-\frac{x^2}{q^2}=6$,where $p,q,m, r$ are constants.
Show that $6(p-qm)(p+qm) \leq r^2 $
From $\frac{y^2}{p^2}-\frac{x^2}{q^2}=6$,
$q^2 y^2 - p^2 x^2 =6p^2 q^2 $ $-(1)$
So, subst. $y=mx+r$ into $(1)$ to get a mess of
$(q^2 m^2 -p^2)x^2 + 2q^2 mrx +r^2 -6p^2 q^2 =0$
For line to meet curve, Discriminant $\geq$ $0$
$b^2 -4ac$ $\geq 0$
So it's
$(2q^2 mr)^2 - 4(q^2 m^2 -p^2)(r^2 -6p^2 q^2) \geq 0$
Continuing from here,
$q^4 m^2 r^2 -q^2 m^2 r^2 + 6p^2 q^4 m^2 +p^2 r^2 -6p^4 q^2 \geq 0$
Then I'm lost
You wrote: $$(q^2 m^2 -p^2)x^2 + 2q^2 mrx +r^2 -6p^2 q^2 =0$$ The mistake in $r^2$.
It should be $r^2q^2$.