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The chords of contact of a poi...

# The chords of contact of a point with respect to a hyperbola and its auxillary circle are perpendicular, then the point lies on :

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One of the directrices one of the asymptotes conjugate hyperbola Axis of hyperbola

A
Solution :
(i) Let theta = tan^(-1)2." "x=sin2theta=(2tantheta)/(1+tan^(2)theta)=(4)/(5) <br> Let alpha=tan^(-1).(4)/(3), " "y=sin.(alpha)/(2)=sqrt((1-cosalpha)/(2))=sqrt((1-3//5)/(2))=sqrt((1)/(5)) <br> (ii) Take tan on both sides ((3)/(4)+(1)/(7))/(1-(3)/(28))=tan.(pi)/(4) <br> (iii) Put x = tan theta. " "therefore 6theta-8theta+4theta=(pi)/(3)impliestheta=(pi)/(6). <br> (because2 thetain "I quadrant"impliessin^(-1)(sin2theta)=2theta,cos^(-1)(cos2theta)=2theta,tan^(-1)(tan2theta)=2theta) <br> (iv) tan^(-1)((1)/(5))lttan^(-1)(sqrt(2)-1)" "therefore0lt2tan^(-1)((1)/(5))lt(pi)/(4)

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Class 12th

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Class 12th

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