Conics Q

**vuze88** · September 5th 2009, 09:58 PM

The tangent at $P(a\sec\theta,b\tan\theta)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has equation $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$ .

If teh tangent at P is also tangent to the circle center $(ae,0)$ and radius $ae\sqrt{e^2+1}$ , show that $\sec\theta=-e$ where e is the eccentricity of the hyperbola.

**Grandad** · September 7th 2009, 04:32 AM

Hello vuze88

Originally Posted by vuze88

The tangent at $P(a\sec\theta,b\tan\theta)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has equation $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$ .

If teh tangent at P is also tangent to the circle center $(ae,0)$ and radius $ae\sqrt{e^2+1}$ , show that $\sec\theta=-e$ where e is the eccentricity of the hyperbola.

Are you sure you haven't made a mistake in the question? I make it that the radius of the circle should be $a\sqrt{e^2+1}$ .

You then solve the problem by saying that a line is a tangent to a circle if and only if the distance from the centre of the circle to the line is equal to the radius of the circle. So in this case we get (using the distance formula $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ with the point $(ae,0)$ and the line $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$ ):

$\frac{|e\sec\theta-1|}{\sqrt{\frac{\sec^2\theta}{a^2}+\frac{\tan^2\th eta}{b^2}}}=a\sqrt{e^2+1}$

$\Rightarrow\frac{a^2b^2(e\sec\theta-1)^2}{b^2\sec^2\theta+a^2\tan^2\theta}=a^2(e^2+1)$

$\Rightarrow\frac{a^2(e^2-1)(e\sec\theta-1)^2}{a^2(e^2-1)\sec^2\theta+a^2(\sec^2\theta-1)}=e^2+1$ , using $b^2 = a^2(e^2-1)$ and $\tan^2\theta=\sec^2\theta -1$

$\Rightarrow\frac{(e^2-1)(e\sec\theta-1)^2}{e^2\sec^2\theta-1}=e^2+1$

$\Rightarrow\frac{(e^2-1)(e\sec\theta-1)^2}{(e\sec\theta-1)(e\sec\theta+1)}=e^2+1$

$\Rightarrow\frac{(e^2-1)(e\sec\theta-1)}{(e\sec\theta+1)}=e^2+1$

$\Rightarrow(e^2-1)(e\sec\theta-1)=(e\sec\theta+1)(e^2+1)$

$\Rightarrow e\sec\theta([e^2-1]-[e^2+1])=[e^2+1]+[e^2-1]$

$\Rightarrow -2e\sec\theta = 2e^2$

$\Rightarrow \sec\theta=-e$

Grandad

**Opalg** · September 7th 2009, 11:22 AM

Originally Posted by vuze88

The tangent at $P(a\sec\theta,b\tan\theta)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has equation $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$ .

If the tangent at P is also tangent to the circle center $(ae,0)$ and radius $ae\sqrt{e^2+1}$ , show that $\sec\theta=-e$ where e is the eccentricity of the hyperbola.

The condition for the line to be tangent to the circle is that the distance from the centre of the circle to the line should be equal to the radius of the circle. But the formula for the distance d from the point $(x_0,y_0)$ to the line $px+qy+r=0$ is $d = \frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}.$

So the distance from $(ae,0)$ to the line $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}-1 = 0$ is $|e\sec\theta-1|\above1pt\displaystyle\sqrt{\frac{\sec^2\theta}{ a^2} + \frac{\tan^2\theta}{b^2}}$ . We want this to be equal to $ae\sqrt{e^2+1}$ . That gives the equation

$|e\sec\theta-1| = ae\sqrt{e^2+1}\sqrt{\frac{b^2\sec^2\theta+a^2\tan^ 2\theta}{a^2b^2}}.$

Simplify the big square root on the right-hand side of that equation as much as you can, using the relations $b^2 = a^2(e^2 - 1)$ and $\sec^2\theta-\tan^2\theta=1$ . Then square both sides of the equation, and solve it for $\sec\theta$ . You should find that there are two solutions, $\sec\theta = 1/e$ and $\sec\theta=-e$ . But the first of those solutions cannot occur, because $|\sec\theta|\geqslant1$ and $1/e<1$ .

**vuze88** · September 8th 2009, 12:14 AM

thanks for the help and sorry about the mistake

**vuze88** · September 8th 2009, 12:24 AM

thanks alot

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