I homogenised the equation of auxiliary circle with the equation of normal and then applied the condition of angle b/w posl 90, but I'm unable to get an expression in the form given
@Chirag_Hegde @Azimuddin_Sheikh @Viraam_Rao
Can someone try this please.
i got this quadratic equation
I got a relation between a,b and ∅. U can connect from there. Relation is something like a²/b² = 1+cosec²∅. May be its not as useful or may be..
What's the answer bro ?
I am getting a very weird answer as 3/(5+2√2). Or is it a integer type @Soumyadeep
@Tushar_Rathore can you show the method i am getting some relation in a,b,theta but unable to proceed further
Ohh, lol. Actually i consider tangent equation so that's why my calculation was easy. @Mayank_Chowdhary bro
When i solved it again what i got is
a⁴+2b⁴ = a⁴tan²∅ + a²b²cot²∅ + 5a²b². But not able to proceed further.
Solving this result will give the answer 1
@Tushar_Rathore but how did you get this ?
Well, i will show but can u give me hint how to proceed that further. Does i have to make a perfect square on rhs or something else @Mayank_Chowdhary
Here , u haven't whole squared a²-b². It will be (a²–b²)² @Mayank_Chowdhary
Oh yes...thanks a lot btw
That's why I was thinking where am I wrong !
what I tried to do is from parametric form of the normal i found distance to the normal from the center which will be fixed as radius and angle are fixed
on solving this what i got is the equ posted in post5