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        If a variable straight  line xcos¤ +ysin¤ = p (p is a constant) is a chord of a hyperbola x^2/a^2 + y^2/b^2 = 1 (b>a), substends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius is:a)    ab/(b-2a)^(1/2)b)    a/ (a-b)^(1/2)c)      ab/ (b^2 - a^2)^(1/2)d)      ab/b (b+a)^(1/2)
2 years ago

## Answers : (1)

Shreyansh Shukla
26 Points
							Equation of pair of straight lines passing through the origin(centre of hyperbola) and points of inresection of the variable chord & hyperbola is:$\frac{x^2}{a^2}-\frac{y^2}{b^2}-(\frac{xcos\alpha+ysin\alpha }{p})^2=0$They are at right angles if coefficient of $x^2$coefficient of $y^2$=0 i.e. $(\frac{1}{a^2}-\frac{(cos\alpha)^2 }{p^2})-(\frac{1}{b^2}+\frac{(sin\alpha )^2}{p^2})=0$ $\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{p^2}$ $p=\frac{ab}{\sqrt{b^2-a^2}}$ As p is length of perpendicular form origin on the line $xcos\alpha +ysin\alpha =p$, the line touches the circle with centre at origin and radius = $\frac{ab}{\sqrt{b^2-a^2}}$

one month ago
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• Mind Map
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• NCERT Solutions
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