Hello UltraGirl Quote:

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**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:$\displaystyle x = ct, y = \frac ct$

From these we get:$\displaystyle \frac{dx}{dt}= c, \frac{dy}{dt} = -\frac c{t^2}$

$\displaystyle \Rightarrow \frac{dy}{dx}= \frac{dy}{dt}\div\frac{dx}{dt}$$\displaystyle = -\frac1{t^2}$

So the equation of the tangent at $\displaystyle \Big(ct, \frac ct\Big)$ is:$\displaystyle y - \frac ct = -\frac1{t^2}\Big(x - ct\Big)$

i.e.$\displaystyle t^2y+x = 2ct$

The intercepts of this line with the coordinate axes are $\displaystyle \Big(0,\frac{2c}{t}\Big)$ and $\displaystyle \Big(2ct,0\Big)$. And if the length of this line segment is $\displaystyle 6c$, then:$\displaystyle (6c)^2 = \Big(\frac{2c}{t}\Big)^2+(2ct)^2$

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

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

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

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

$\displaystyle \Rightarrow t = \pm\sqrt{\frac{9\pm\sqrt{77}}{2}}$

Thus we have found four values of $\displaystyle t$, which give the four required points when substituted into the parametric co-ordinates $\displaystyle \Big(ct, \frac ct\Big)$.

Grandad