Math Help - [SOLVED] Hyperbola

1. [SOLVED] Hyperbola

Hello.

For this question - Show how to determine a point P of the hyperbola xy = c^2 such that the asymptotes cut off a segment of length 6c in the tangent at P,

I've equated 6c^2= sqrt[a^2+b^2], with the x intercept being (a,0) and the y intercept being (0,b) BUT that gets me nowhere cause I now have 3 variables to work with.

I tried finding the equation of the tangent, which is y = b/a * x + b.

Please could someone show me how to do this?

2. Hello UltraGirl
Originally Posted by UltraGirl
Hello.

For this question - Show how to determine a point P of the hyperbola xy = c^2 such that the asymptotes cut off a segment of length 6c in the tangent at P,

I've equated 6c^2= sqrt[a^2+b^2], with the x intercept being (a,0) and the y intercept being (0,b) BUT that gets me nowhere cause I now have 3 variables to work with.

I tried finding the equation of the tangent, which is y = b/a * x + b.

Please could someone show me how to do this?
I assume that you know how to use the parametric equations of the rectangular hyperbola:
$x = ct, y = \frac ct$
From these we get:
$\frac{dx}{dt}= c, \frac{dy}{dt} = -\frac c{t^2}$

$\Rightarrow \frac{dy}{dx}= \frac{dy}{dt}\div\frac{dx}{dt}$
$= -\frac1{t^2}$
So the equation of the tangent at $\Big(ct, \frac ct\Big)$ is:
$y - \frac ct = -\frac1{t^2}\Big(x - ct\Big)$
i.e.
$t^2y+x = 2ct$
The intercepts of this line with the coordinate axes are $\Big(0,\frac{2c}{t}\Big)$ and $\Big(2ct,0\Big)$. And if the length of this line segment is $6c$, then:
$(6c)^2 = \Big(\frac{2c}{t}\Big)^2+(2ct)^2$

$\Rightarrow 36c^2 = \frac{4c^2}{t^2}+4c^2t^2$

$\Rightarrow 9t^2 = 1 +t^4$

$\Rightarrow (t^2)^2-9t^2+1=0$

$\Rightarrow t^2 = \frac{9\pm\sqrt{81-4}}{2}$

$\Rightarrow t = \pm\sqrt{\frac{9\pm\sqrt{77}}{2}}$
Thus we have found four values of $t$, which give the four required points when substituted into the parametric co-ordinates $\Big(ct, \frac ct\Big)$.