@Azimuddin_Sheikh @Chirag_Hegde @pratyaksh_tyagi @Siddhant_Mudholkar @Mayank_Chowdhary
Guys, I solved the question by writing the distance AB as a function of sin ϴ and then maximising it.
But it was a very long solution.
Can anyone think of a simpler and shorter way to do this?
What's the answer given ?
Getting 2 by long method @Azimuddin_Sheikh
For short method try using the common tangency condition that slope must be +- √(K-b^2)/(a^2-k) , here k is square of radius of circle
Didn’t get u bro
Bro that slope tells us about the common tangents to both ellipse and a given circle . u just find one coordinate in ellipse or circle part and then solve for maximizing distance easily from this condition
I did the same I guess.
I took a parametric point on ellipse and found the intersection point of tangent at that point with circle. Then maximized distance.
It’s too long.
are u talking about this method or something else ?
Oh yeah just a less bit calculation for other point part as such we know the slope . I will think about another method after sometime then
Ok bro. Please try for another method if possible
Got a short method
Yay! Please post.
How did you get AB?
I kind of didn't get it
Distance between parallel lines N and N’
N should have a zero not 12 on RHS though
Yup by mistake
Ah got it
Nice way @Anish_Saparia