IIT JEE , AIEEE Forum - Mathematics , Vectors

abhishek_varshney

show that the normal at the point (3t,4/t) of the curve xy=12 cuts the curve again at the point whose parameter t1 is given by t1=(-16/9t^3).

bsgdabest

given xy=12
u gave parametric pt as (3t,4/t)

derive both sides, and get dy/dx at (t) = -4/3t^2 so normal slope at t= 3t^2/4
i am writing normal equation: y-4/t=3t^2(x-3t)/4
since it cuts the hyperbola again , its like we r solving it with xy=12, put y=12/x
12/x-4/t=3t^2(x-3t)/4
finally we get 3t^3x^2+x(16-9t^3)-48t=0
it needs to have 2 roots, and their product, say at x₂ (let the pt(t1) be(x1,y1) ]
x₁x₂₌-48/t^2
ur x₁=3t and say x₂=3t₂., we get t₂=-16/9t^3

Ask & Discuss Questions with Community & Experts