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the angle between a pair of tangents from a point p to the circle x2 +y2+4x-6y+9sin2@+13cos2@=0 is 2@.find the equation of locus of point p.

ishita chauhan , 5 Years ago
Grade 12th pass
anser 1 Answers
Vikas TU

Last Activity: 5 Years ago

Dear student 
S = x2 +y2+4x-6y+9sin2@+13cos2@=0
C(-2,3)
r = Sqr((-2)^2 + 3^2 - 9Sin^2@ - 13Cos^2@)
=> sqrt (13- 9Sin^2@-13Cos^2@)
LT = sqrt(h^2+k^2+4h -6k +9sin^2@+13cos^2@)
tan@ = 2Sin@/sqrt(h^2+k^2+4h -6k +9sin^2@+13cos^2@)
sqrt(h^2+k^2+4h -6k +9sin^2@+13cos^2@) = 2Sin@ / Sin@ *Cos@
(h^2+k^2+4h -6k +9sin^2@+13cos^2@) = 4Cos^2@
(x^2+y^2+4x -6y +9sin^2@+9cos^2@) =0
x^2+y^2+4x -6y +9 =0 Locus 

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