Question on Ellipse, tried about ten times....

"The portion of ellipse (take the standard one) in first quadrant is divided by two lines y=m1x and y=m2x into three equal parts. Then m1*m2=???"

I thought that i would do like the projection of auxilary circle is ellipse,, or second i did was the ratio of ellipse's area and auxillary circle's area is b:a (as pi*a*b:pi*a*a) after solving i got the answer but i what i got was eccentric angles not the angles at which they are inclined at????


Anyways answer is 1/4...........

Area of ellipse sector OP₁P₂ where P₁, P₂ have eccentric angles θ₁, θ₂ is ½ab(θ₂−θ₁)

(this result can be derived geometrically without calculus)


Gradient of OP = m = (bsinθ)/(acosθ) = (b/a)tanθ → θ = tan⁻¹(am/b) in Q1

∴ area of sector OP₁P₂ = ½ab{ tan⁻¹(am₂/b) − tan⁻¹(am₁/b) }


If three sectors formed by m=0, m=m₁, m=m₂ and m=∞ are equal …

tan⁻¹(am₁ /b)−0 = tan⁻¹(am₂/b)−tan⁻¹(am₁/b) = π/2−tan⁻¹(am₂/b)

→ tan⁻¹(am₁ /b)=π/6, tan⁻¹(am₂/b)=π/3

→ m₁ = (b/a)(1/√3), m₂=(b/a)√3 → m₁m₂ = b²/a²


Does b/a have a specific value for your problem ?

angle made by lines with positive direction of x axis = 30degree , 60 degree respectively
their slopes are = tan30 , tan 60 respectively
so m1.m2 = tan30,tan60 = 1/sqrt3 * sqrt3 = 1
........................answer