Questions
Let be a variable point such that
Statement-1: traces a hyperbola whose eccentricity is.
Statement-2: traces a hyperbola such that the equation of its conjugate axis is
detailed solution
Correct option is B
Distance of P from the fixed points F1(3, 2) and F2(6,−2) is constant 3.So by definition, locus of P is a hyperbola with foci F1 and F2, length of the transverse axes is 3. If e is the eccentricity then F1F2=2ae where 2a=3. ⇒e=139+16=53 and the statement-1 is true.n statement-2 slope of the conjugate axis is −6−3−2−2=34 and it passes through the mid-point of F1F2 So its equation isy=34x−92⇒6x−8y=27Thus statement-2 is also true but does not lead to statement-1.Talk to our academic expert!
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