conic question please help!!!! (1 Viewer)

coeyz · Oct 25, 2009

P (ct, c/t) lies on xy=c^2 the normal of P meets the hyperbola again at Q. PQ is a diameter of a circle, and the circle meets R again. Find the coordinates of Q and R. THANKS A LOT!

kwabon · Oct 25, 2009

coeyz said:
P (ct, c/t) lies on xy=c^2 the normal of P meets the hyperbola again at Q. PQ is a diameter of a circle, and the circle meets R again. Find the coordinates of Q and R. THANKS A LOT!

sorry, there is something really fucked up with this site atm, something is wrong with my post, cant upload or copy any pictures into it, and "enter" does not work, as i cannot go onto next line, so i wont be able to show any working but will guide u through the steps. derive xy = c^2 and find the gradient and equation of the normal. simultaneously solve the normal with xy = c^2 to give u P. the midpoint of P and Q is the center of the circle. use the distance formula to work out the distance of the diameter. hence finding the equation of circle. simultaneously solve xy = c^2 with the equation of the circle to obtain R. hope this helps =)

jet · Oct 25, 2009

$\text{Differentiating the equation of the hyperbola, } \\ y + x\frac{dy}{dx} = 0 \\ \frac{dy}{dx} = -\frac{y}{x} \\ \therefore \text{at } P, \frac{dy}{dx} = -\frac{1}{t^2} \\ \therefore M_{normal} = t^2 \text{ since } -\frac{1}{t^2} \cdot t^2 = -1 \\ y - \frac{c}{t} = t^2 \left (x - ct \right ) \\ t^3x - ty = c\left (t^4 - 1 \right ) \\ \text{Solving this with } xy = c^2, t^3x - \frac{c^2t}{x} = c\left (t^4 - 1 \right ) \\ t^3x^2 - c\left (t^4 - 1 \right )x - c^2t = 0 \text{one of these roots will be } x = ct \\ \text{ and the other will be our root of interest. We can just use the sum of roots.} \\ ct + x_0 = \frac{c\left (t^4 - 1 \right )}{t^3} \\ x_0 = \frac{c\left (t^4 - 1 \right )}{t^3} - ct \\ = -\frac{c}{t^3} \\ \therefore y_0 = -ct^3 \\ \therefore Q \equiv \left (-\frac{c}{t^3}, -ct^3 \right )$

$\text{Now, the centre of the circle is the midpoint of } PQ \\ \therefore \text{The coordinates of C are } C \equiv \left ( \frac{ct - \frac{c}{t^3}}{2},\frac{\frac{c}{t} - ct^3}{2} \right ) \\ = \left (\frac{c \left (\frac{t^4 - 1}{t^3} \right )}{2}, \frac{c\left (\frac{1 - t^4}{t} \right )}{2} \right ) \\ = \left (\frac{c\left (t^4 - 1 \right )}{2t^3}, \frac{c\left (1 - t^4 \right )}{2t} \right ) \\ \text{Just to simplify matters, the radius of the circle is just half of } PQ \\ \therefore PQ^2 = \left (ct + \frac{c}{t^3} \right )^2 + \left (\frac{c}{t} + ct^3 \right )^2 \\ = \left (\frac{c\left (t^4 + 1 \right )}{t^3} \right )^2 + \left (\frac{c\left (t^4 + 1 \right )}{t} \right)^2 \\ = \left (c^2 \left (t^4 + 1 \right )^2 \right )\left (\frac{1}{t^6} + \frac{1}{t^2} \right ) \\ = \left (c^2 \left (t^4 + 1 \right )^2 \right )\left (\frac{1 + t^4}{t^6} \right ) \\ = \frac{c^2 \left (t^4 + 1 \right )^3}{t^6} \\$

$\text{Now, the circle will have equation } \left (x - \frac{c\left (t^4 - 1 \right )}{2t^3} \right )^2 + \left (y - \frac{c\left (1 - t^4 \right )}{2t} \right )^2 = \left (\frac{PQ}{2} \right )^2 = \frac{PQ^2}{4} \\\therefore \left (x - \frac{c\left (t^4 - 1 \right )}{2t^3} \right )^2 + \left (y - \frac{c\left (1 - t^4 \right )}{2t} \right )^2 = \frac{c^2 \left (t^4 + 1 \right )^3}{4t^6} \\ \text{We solve this with } xy = c^2 \\ \left (x - \frac{c\left (t^4 - 1 \right )}{2t^3} \right )^2 + \left (\frac{c^2}{x} - \frac{c\left (1 - t^4 \right )}{2t} \right )^2 = \frac{c^2 \left (t^4 + 1 \right )^3}{4t^6} \\ \left (x - \frac{c\left (t^4 - 1 \right )}{2t^3} \right )^2 + \frac{1}{x^2}\left (c^2 - \frac{c\left (1 - t^4 \right )x}{2t} \right )^2 = \frac{c^2 \left (t^4 + 1 \right )^3}{4t^6} \\ x^2\left (x - \frac{c\left (t^4 - 1 \right )}{2t^3} \right )^2 - \frac{c^2 \left (t^4 + 1 \right )^3x^2}{4t^6} + \left (c^2 - \frac{c\left (1 - t^4 \right )x}{2t} \right )^2 = 0$

$\text{There are 4 roots to this equation. One is the point Q. } \\ \text{Now considering the point P. The line PQ is the normal at P, and also the diameter } \\ \text{of the circle (thus it is also the radius.)} \\ \text{Since the normal is perpendicular to the tangent of the hyperbola,} \\ \ \text{AND the radius of a circle is perpendicular to the tangent of the circle,} \\ \text{THEN, the circle and the hyperbola must share a tangent.} \\ \text{As a result, there is a double root of this equation at P.} \\ \text{We can use the sum of roots to get the point R} \\ \text{The coefficient of the } x^4 \text{ term by inspection is } 1 \\ \text{The coefficient of the } x^3 \text{ term, by inspection is } -\frac{c\left (t^4 - 1 \right )}{t^3} \\$

$\text{Thus, the sum of roots is } ct + ct - \frac{c}{t^3} + x_R = \frac{c\left (t^4 - 1 \right )}{t^3} \\ x_R + \left (\frac{c\left (2t^4 - 1 \right )}{t^3} \right ) = \frac{c\left (t^4 - 1 \right )}{t^3} \\ \therefore x_R = \frac{c\left (t^4 - 1 \right )}{t^3} - \left (\frac{c\left (2t^4 - 1 \right )}{t^3} \right ) \\ = \frac{-c t^4}{t^3} \\ = -ct \\ y_R = \frac{c^2}{x_R} \\ = \frac{c^2}{-ct} \\ = -\frac{c}{t} \\ \therefore Q \equiv \left (-\frac{c}{t^3}, -ct^3 \right ), R \equiv \left (-ct, -\frac{c}{t} \right )$

I think I'm done helping people for the night, lol.

Aerath · Oct 26, 2009

coeyz said:
P (ct, c/t) lies on xy=c^2 the normal of P meets the hyperbola again at Q. PQ is a diameter of a circle, and the circle meets R again. Find the coordinates of Q and R. THANKS A LOT!

I'm not quite sure where "R" is. Do you mean it's where the circle intersects with the hyperbola again?

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conic question please help!!!! (1 Viewer)

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