Question

Find points p and q on the parabola y=1-x^2 so that the triangle abc formed by the x-axis and the tangent lines at p and q is an equilateral triangle.

Answer 1

from your question, daigram is drawn as shown in below attachment.

first of all, we assumed co-ordinate of equilateral triangle ABC .
B = (-c , 0) , C = (c , 0) and A = (0, c√3)
so, slope of line AB = (c√3 - 0)/(0 +c) = √3...(1)
slope of line AC = (c√3 - 0)/(0 - c) = -√3....(2)
now , equation of parabola is y = 1 - x²
differentiate it wrt to x
y' = -2x ,
Let point P is (a,b)
then, slope of line AB = -2a = √3 [ from equation (1), ]
a = -√3/2
b = 1 - a² [ as it is on parabola ]
b = 1 - (-√3/2)² = 1/4

Similarly, slope of line AC
y' = -2x ,
Let point Q is (a',b')
then, slope of line AC = -2a' = -√3
a' = √3/2
b' = 1 - a'² = 1 - (√3/2)² = 1/4

hence, point P(-√3/2,1/4) and (√3/2,1/4)