Paragraph: Consider a conic C: y^2=4 x . Let P Q R be an equilateral triangle with side length k where P be any point on C, Q be the foot of perpendicular from P upon the directrix of C and R be the focus of C . A circle C₂ is inscribed in another conic C₁: y^2=k(x+1) which touches C₁ at the points where C₁ cuts the y -axis. C₃ is an ellipse whose auxiliary circle is C₂ and major axis coincides with the axis of symme

Step-by-step solution

We have P M=1+t^2 aligned & P S= (t^2-1 )^2+4 t^2 = (t^2+1 ) & M S= 4+4 t^2 =2 1+t^2 & 2 1+t^2 =1+t^2 aligned P M=1+t^2=4=a=k (Given) Hence C₁: y^2=4(x+1) Equation of tangent to C₁ at (0,2) is aligned & 2 y=4 ( x+0 2 +1 ) & y=x+2 aligned Now circle which touches above line at (0,2) , is x^2+(y-2)^2+ (x-y+2)=0 . As above circle is passing through the point (0,-2) , so aligned & 0+16+ (4)=0 =-4 & C₂: x^2+(y-2)^2-4(x-y+2)=0 & or, C₂: x^2+y^2-4 x-4=0 aligned Now C₃: (x-2)^2 a ^2 + y ^2 ~b ^2 =1, a =2 2 and b =2 So C ₃: