- #1
ManInTheSuit97
- 1
- 0
Minimum Distance between y^2=4x and x^2+y^2-12x+31=0.
Attempt:I got that the parabola has vertex at(0,0) and focus at(1,0).The Circle is centred at (6,0)and its radius is sqrt 5.I figured that the double ordinate that passed through (6,0) would be bisected at the point.So I found out the chord of contact and it turned out to be x=6.I substituted the value of x in the parabola and found out y= sqrt24.I thought maybe the minimum distance is the difference in vertical distance and so my answer was sqrt 24-sqrt5.But it is not the answer and so I probably have messed up somewhere.I want to know the approach I should take
Attempt:I got that the parabola has vertex at(0,0) and focus at(1,0).The Circle is centred at (6,0)and its radius is sqrt 5.I figured that the double ordinate that passed through (6,0) would be bisected at the point.So I found out the chord of contact and it turned out to be x=6.I substituted the value of x in the parabola and found out y= sqrt24.I thought maybe the minimum distance is the difference in vertical distance and so my answer was sqrt 24-sqrt5.But it is not the answer and so I probably have messed up somewhere.I want to know the approach I should take