Let P be the point on the parabola y²=4x which is at the shortest distance from the center S of the circle x²+y²-8x-6y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then SP=2√5. The x-intercept of the normal to the parabola at P is 6. SQ:QP=(√5+1):2. The slope of the tangent to the circle at Q is 12.

Let any point P(t²,2t) on parabola. As we know shortest distance between two curves lies along their common normal. The common normal will pass through centre of circle. Slope of normal to the parabola y²=4x at P(t,2t) is -t. ⇒ t³=8 ⇒ t=2 ∴ P(4,4) ∴ SP=√(4-4)²+(4-4)²=2√5 (i) equation of normal at P(4,4) ⇒ y-4=-1/2(x-4) ⇒ y=-x/2+6 ⇒ x-intercept=6 (ii) slope of tangent at Q = slope of tangent at P = 1/2 (iii) SQ/PQ = 2/2√5 = 1/√5 = (√5+1)/4