# Thread: Tangent of the hyperbola

1. ## Tangent of the hyperbola

I'd like some help with this rather difficult question:

The tangents at $\
P(ct_1 ,\frac{c}{{t_1 }})
\$
and $\
Q(ct_2 ,\frac{c}{{t_2 }})
\$
to the rectangular hyperbola with equation $\
xy = c^2
\$
meet on the rectangular hyperbola with equation $\
xy = \frac{{c^2 }}{4}
\$
. Prove that PQ is tangent to the curve with equation $\
xy = 4c^2
\$

I'm not really sure what I need to do to prove this statement, and I can't seem to find the equation of PQ. Any help would be greatly appreciated.

2. Originally Posted by free_to_fly
I'd like some help with this rather difficult question:

The tangents at $\
P(ct_1 ,\frac{c}{{t_1 }})
\$
and $\
Q(ct_2 ,\frac{c}{{t_2 }})
\$
to the rectangular hyperbola with equation $\
xy = c^2
\$
meet on the rectangular hyperbola with equation $\
xy = \frac{{c^2 }}{4}
\$
. Prove that PQ is tangent to the curve with equation $\
xy = 4c^2
\$

I'm not really sure what I need to do to prove this statement, and I can't seem to find the equation of PQ. Any help would be greatly appreciated.
The good news is I've solved your problem.

The bad news is the solution is horrible.

Stay tuned ......

3. Originally Posted by mr fantastic
The good news is I've solved your problem.

The bad news is the solution is horrible.

Stay tuned ......
Equation of tangent:

$y = \frac{c^2}{x} \Rightarrow \frac{dy}{dx} = - \frac{c^2}{x^2}$. So at x = ct, $m = - \frac{1}{t^2}$.

Using $y - y_1 = m(x - x_1)$:

At P: $y = -\frac{1}{t_1^2} x + \frac{2c}{t_1}$.

At Q: $y = -\frac{1}{t_2^2} x + \frac{2c}{t_2}$.

Intersection point of these two tangents:

$x = \frac{2ct_1 t_2}{t_1 + t_2} \,$ and $y = \frac{2c}{t_1 + t_2}$.

But this intersection point lies on the hyperbola $xy = \frac{c^2}{4}$. Sub them in:

$\left( \frac{2ct_1 t_2}{t_1 + t_2} \right) \left( \frac{2c}{t_1 + t_2}\right) = \frac{c^2}{4}$

$\Rightarrow t_1 t_2 = \frac{(t_1 + t_2)^2}{16}$ .... (A)

Equation of line PQ:

$m = -\frac{1}{t_1 t_2}$.

Using $y - y_1 = m(x - x_1)$:

$y = -\frac{1}{t_1 t_2} x + \frac{c(t_1 + t_2)}{t_1 t_2}$.

Substitute equation (A):

$y = - \frac{16}{(t_1 + t_2)^2} x + \frac{16c}{t_1 + t_2}$.

Tangent to curve $xy = 4c^2\,$ at $\left( 2ct_3, \, \frac{2c}{t_3} \right)$:

Using the same approach as before (or just make the replacement c --> 2c in the other tangents), you get:

$y = -\frac{1}{t_3^2} x + \frac{4c}{t_3}$.

So ...... is there a value of $t_3$ that gives $y = - \frac{16}{(t_1 + t_2)^2} x + \frac{16c}{t_1 + t_2} \,$, the line PQ?

Compare the tangent and the line:

Gradient (coefficient of x): $\frac{1}{t_3^2} = \frac{16}{(t_1 + t_2)^2} \Rightarrow t_3 = \frac{t_1 + t_2}{4}$.

Does this value satisfy $\frac{4c}{t_3} = \frac{16c}{t_1 + t_2}$ ....? Obviously.

So there you go. The line PQ is tangent to $xy = 4c^2\,$, at the point $\left( 2ct_3, \, \frac{2c}{t_3} \right)$ where $t_3 = \frac{t_1 + t_2}{4}$.

4. Lol I just knew it couldn't have been a very pleasant, short and neat answer, seeing how it's towards the end of the book, but if it's hard it'll make me feel better about not being able to do it.

5. Originally Posted by free_to_fly
Lol I just knew it couldn't have been a very pleasant, short and neat answer, seeing how it's towards the end of the book, but if it's hard it'll make me feel better about not being able to do it.
Yeah, well watch this space ..... Someone's sure to give a simple obvious proof now, and have a good laugh at all this ....