1. ## Hyperbola and tangents

Hi all, this is a really long problem, so please bear with me, parts a - f is done and solution is provided, so it's only g that I need help on, but I figured a - f would help... and once again, any help that you can offer is great!

Here's the question:

The line $\displaystyle l$ is a common tangent to two hyperbolas: $\displaystyle xy=c^2$ and $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with points of contact $\displaystyle P$ and $\displaystyle Q$ respectively.

(Unfortunately I don't know how to do diagrams on here, so you may have to draw this one out)

a) Show that $\displaystyle l$ has equation $\displaystyle x+t^2y=2ct$. - DONE!

b) Show that $\displaystyle l$ also has equation $\displaystyle \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$. - DONE!

c) Deduce that $\displaystyle \frac{\sec\theta}{a}=\frac{-\tan\theta}{bt^2}=\frac{1}{2ct}$. - DONE!

d) Write the coordinates of $\displaystyle Q$ in terms of $\displaystyle t$, $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$, and show that $\displaystyle b^2t^4+4c^2t^2-a^2=0$. Deduce that there are exactly two such common tangents to the hyperbola. - DONE!

$\displaystyle Q = (a\sec\theta,b\tan\theta)=(\frac{a^2}{2ct},\frac{-b^2t}{2c})$

Treating $\displaystyle b^2t^4+4c^2t^2-a^2=0$ as a quadratic equation in $\displaystyle t^2$, and consider its discriminant, one can show that there are two common tangents to the hyperbolas.

e) Using the summetry in the graphs to draw in the second common tangent with points of contact $\displaystyle R$ on $\displaystyle xy=c^2$ and $\displaystyle S$ on $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Write the coordinates of $\displaystyle R$ and $\displaystyle S$ in terms of $\displaystyle t$, $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$. - DONE!

$\displaystyle R=(-ct,-\frac{c}{t})$
$\displaystyle S=(-\frac{a^2}{2ct}, \frac{b^2t}{2c})$

f) Show that if $\displaystyle PQRS$ is a rhombus, then $\displaystyle b^2=a^2$ and deduce that $\displaystyle t^2<1$ - DONE!

To show $\displaystyle b^2=a^2$, simply solve $\displaystyle d_{PQ}=d_{QR}$

$\displaystyle b^2t^4+4c^2t^2-a^2=0$
$\displaystyle a^2t^4+4c^2t^2-a^2=0$
$\displaystyle a^2t^4-a^2=-4c^2t^2$
$\displaystyle a^2(t^4-1)=-4c^2t^2$

The right hand side is strictly negative (as $\displaystyle c\ne0$ and $\displaystyle t\ne0$), therefore the left side must also be negative.

$\displaystyle a^2(t^4-1)<0$
$\displaystyle (t^4-1)<0$
$\displaystyle (t^2+1)(t^2-1)<0$
$\displaystyle (t^2-1)<0$
$\displaystyle t^2<1$

g) Show that if $\displaystyle PQRS$ is a square, then $\displaystyle t^4+2t^2-1=0$ and deduce that $\displaystyle 2c^2=a^2$. What is the relationship between the two hyperbolas if $\displaystyle PQRS$ is a square?

NEED HELP: I can't show the first part... but the deduction I can do once the first part is done... and then I haven't yet attempted the relationship (but I'm sure I'll have no trouble doing that)

THANKS A LOT GUYS, IT'LL REALLY HELP ME OUT IF YOU COULD GIVE ME A HAND WITH PART g)

2. Hello AlvinCY
Originally Posted by AlvinCY
...

g) Show that if $\displaystyle PQRS$ is a square, then $\displaystyle t^4+2t^2-1=0$ and deduce that $\displaystyle 2c^2=a^2$. ...
Gradient PQ = $\displaystyle -\frac{1}{t^2}$

Gradient QR = $\displaystyle \frac{-\dfrac{b^2t}{2c}+\dfrac{c}{t}}{\dfrac{a^2}{2ct}+ct }=\frac{-b^2t^2+2c^2}{a^2+2c^2t^2}$

PQRS is a square $\displaystyle \Rightarrow PQ\perp QR$

$\displaystyle \Rightarrow \Big(-\frac{1}{t^2}\Big).\Big(\frac{-b^2t^2+2c^2}{a^2+2c^2t^2}\Big)=-1$

$\displaystyle \Rightarrow -b^2t^2+2c^2 = a^2t^2+2c^2t^4$

$\displaystyle \Rightarrow 2c^2t^4+(a^2+b^2)t^2-2c^2=0$

$\displaystyle \Rightarrow 2c^2t^4+2a^2t^2-2c^2=0$ (since $\displaystyle b^2=a^2$) (1)

Now $\displaystyle t$ also satisfies $\displaystyle a^2t^4+4c^2t^2 -a^2 = 0$, using (d), and again using $\displaystyle b^2=a^2$

Subtract this equation from (1):

$\displaystyle (2c^2-a^2)t^4-2(2c^2-a^2)t^2-(2c^2-a^2) = 0$

$\displaystyle \Rightarrow 2c^2-a^2 = 0$ or, if not, $\displaystyle t^4-2t^2-1 = 0$, by dividing through by the common factor.

Now $\displaystyle t^2 < 1 \Rightarrow t^4-1<0$. Hence $\displaystyle t^4-1 \ne 2t^2$ for real values of $\displaystyle t$. So $\displaystyle t^4-2t^2-1 \ne 0$.

Hence $\displaystyle 2c^2 - a^2 = 0$ and so, eliminating $\displaystyle a$ in equation (1):

$\displaystyle \Rightarrow 2c^2t^4+c^2t^2-2c^2=0$

$\displaystyle \Rightarrow t^4 +2t^2 -1 = 0$

I seem to have proved this in the reverse order, so I suspect there may be another method. However, I think the above reasoning is quite sound!

I did try the mulitplication of the two gradients, but didn't see past the step where you used (d) again! That was marvellous though, I must say. Thank you again!