conic sections
If a variable chord of hyperbola $x^2$$/4$ – $y^2$$/8$ $=$ $1$ subtends a right angle at the centre of hyperbola . If the chord touches a fixed concentric circle with hyperbola then we have to find the radius of the circle .
I thought of doing it by homozenizing , but not able to do how ?
Viewing a vertical hyperbola in a direction perpendicular to the cutting plane, the "edges" of the cone appear as the hyperbola's asymptotes.
Consequently, the diameter of the circle in question is simply how "wide" those asymptotes are at the level of the hyperbola's vertex. That is, if $A$ is a vertex of the hyperbola with center $O$, and $A^\prime$ is the point where the perpendicular to $\overline{OA}$ at $A$ meets an asymptote, then $|AA^\prime|$ is the radius of the circle. This distance is precisely the length of the conjugate semi-axis.
Writing $a$, $b$, $c$ for the transverse semi-axis, the conjugate semi-axis, and the distance from center to focus, and writing $e$ for the hyperbola's eccentricity, we know that $$a^2 + b^2 = c^2 \quad\to\quad b^2 = c^2 - a^2 = a^2\left(\frac{c^2}{a^2}-1\right) = a^2 \left(e^2-1\right)$$
So, the horizontal circle that meets a vertical hyperbola at its vertex has radius $b = a \sqrt{e^2-1}$. $\square$
The line equation seems a bit of a distraction. Here's how I'd solve the problem.
Suppose $H$ and $K$ determine a chord of the hyperbola that subtends a right angle at its center (the origin), $O$. We can write $$H = (h \cos\phi, h \sin\phi) \qquad K = (k \sin\phi,-k \cos\phi)$$ where $h:=|OH|$, $k:=|OK|$, and $\phi$ is some arbitrary angle. Define $p := |OP|$, where $P$ is the foot of the altitude from $O$ to $\overleftrightarrow{HK}$. Calculating the area of $\triangle HOK$ in two ways, we have
$$\frac12|OH||OK| = \frac12|OP||HK|\quad\to\quad hk = p\sqrt{h^2+k^2}\quad\to\quad p = \frac{1}{\sqrt{\dfrac{1}{h^2}+\dfrac{1}{k^2}}} \tag{1}$$
Since $H$ and $K$ both lie on the hyperbola, we have $$\begin{align} \frac{h^2}{a^2}\cos^2\phi - \frac{h^2}{b^2}\sin^2\phi &= 1 \tag{2} \\[4pt] \frac{k^2}{a^2}\sin^2\phi - \frac{k^2}{b^2}\cos^2\phi &= 1 \tag{3} \end{align}$$ Therefore, $k^2 (2) + h^2 (3)$ becomes $$\frac{h^2k^2}{a^2}(\cos^2\phi+\sin^2\phi)-\frac{h^2k^2}{b^2}(\sin^2\phi+\cos^2\phi) = h^2 + k^2 \quad\to\quad \frac{1}{a^2}-\frac{1}{b^2} = \frac{1}{h^2} + \frac{1}{k^2} \tag{4} $$ Thus, recalling $(1)$, we have $$p = \frac{1}{\sqrt{\dfrac{1}{a^2}-\dfrac{1}{b^2}}} = \frac{ab}{\sqrt{b^2-a^2}} \tag{5}$$ Since $p$ is the distance from $\overleftrightarrow{HK}$ to the origin, we conclude that the line is always tangent to the circle with the radius given in $(5)$, as desired. $\square$
Best Answer
Homogenise it and then coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.