Conics Help1!!!! (1 Viewer)

kevda1st · Feb 10, 2009

1. The point p(ct , c/t ) lies on the rectangular hyperbolda xy=c^2
The normal at P meets the hyperbola again at Q.
M is the mid point of PQ. FInd the equation of the locus of M.

2. Find the equation of the four tangents common to the hyperbola x^2 - 2y^2 =4 and the circle x+ y^2 = 1.
Find the point of contacne of these tangents with the circle.

Thanks in advanced~

Trebla · Feb 10, 2009

$$Equation of normal at $P\\y - \dfrac{c}{t} = t^2(x-ct)\\$Sub $y = \dfrac{c^2}{x}\\\dfrac{c^2}{x}-\dfrac{c}{t} = t^2(x-ct)\Rightarrow c^2t-cx=x^2t^3-xct^4\\t^3x^2-x(ct^4-c)-c^2t=0\\$The solutions to this equation are $x_1$ and $x_2$ which are the $x$ coordinates of points $P$ and $Q$ respectively$\\$Sum of roots $ x_1 + x_2 = \dfrac{ct^4-c}{t^3}\\$Product of roots $x_1x_2=-\dfrac{c^2}{t^2}\\$The $x$ coordinates of the midpoint $M$ of $P$ and $Q:\\x=\dfrac{x_1+x_2}{2}=\dfrac{ct^4-c}{2t^3}\\$Let $y$ coordinates of $P$ and $Q$ be $y_1$ and $y_2$ respectively$\\\dfrac{y_1+y_2}{2}=\dfrac{\frac{c^2}{x_1}+\frac{c^2}{x_2}}{2}}=\dfrac{c^2(x_1+x_2)}{2x_1x_2}=\dfrac{c^2x}{x_1x_2}=-t^2x\Rightarrow t^2=-\dfrac{y}{x}\\\therefore x^2=\dfrac{c^2(t^4-1)^2}{4t^6}=-\dfrac{c^2(\frac{y^2}{x^2}-1)^2}{4\frac{y^3}{x^3}}=-\dfrac{c^2\frac{(y^2-x^2)^2}{x^4}}{4\frac{y^3}{x^3}}=-\dfrac{c^2(y^2-x^2)^2}{4xy^3}$

Check working...looks weird

jet · Feb 11, 2009

Yeah. I graphed the equations and they dont seem to work with the locus you have found... Though the working seems correct to me. You took the steps I would.

To the OP, what book did you get it out of? There may already be answers out there we can direct you to.

Conics Help1!!!! (1 Viewer)

kevda1st

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Trebla

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jet

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