A variable chord of the hyperbola subtends a right angle at the centre of the hyperbola. If this chord touches a fixed circle which is concentric with the hyperbola and r is radius of the circle, then r2 is

Question

A variable chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{8} = 1$ subtends a right angle at the centre of the hyperbola. If this chord touches a fixed circle which is concentric with the hyperbola and $r$ is radius of the circle, then $r^{2}$ is

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$Let the variable chord be x \cos α + y \sin α = p which intersect the hyperbola in A and B.$

$r^{2}$ $\frac{x^{2}}{4} - \frac{y^{2}}{8} = {(\frac{x \cos α + y \sin α}{p})}^{2}$

$\Rightarrow (\frac{1}{4} - \frac{\cos^{2} α}{p^{2}}) x^{2} - (\frac{1}{8} + \frac{\sin^{2} α}{p^{2}}) y^{2} - (\frac{2 \sin α \cos α}{p^{2}}) x y = 0$

$\Rightarrow \frac{1}{4} - \frac{\cos^{2} α}{p^{2}} - \frac{1}{8} - \frac{\sin^{2} α}{p^{2}} = 0$ $\Rightarrow p^{2} = 8$

$The variable line touches the fixed circle, thus perpendicular distance of (0, 0) = Radius$

$r =$ $| \frac{0 + 0 - p}{\sqrt{\cos^{2} α + \sin^{2} α}} |$

$\Rightarrow r =$ $| p |$ $= \sqrt{8}$

$∴$ $equation of the circle is x^{2} + y^{2} = 8$

Solution

Q.
A variable chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{8} = 1$ subtends a right angle at the centre of the hyperbola. If this chord touches a fixed circle which is concentric with the hyperbola and $r$ is radius of the circle, then $r^{2}$ is

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Solution

Q. A variable chord of the hyperbola x24-y28=1 subtends a right angle at the centre of the hyperbola. If this chord touches a fixed circle which is concentric with the hyperbola and r is radius of the circle, then r2 is

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Q.
A variable chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{8} = 1$ subtends a right angle at the centre of the hyperbola. If this chord touches a fixed circle which is concentric with the hyperbola and $r$ is radius of the circle, then $r^{2}$ is