Let the shortest distance from (a,0), a>0, to the parabola y2=4x be 4. Then the equation of the circle passing through the point (a,0) and the focus of the parabola, and having its centre on the axis of the parabola is :
- x2+y2-8x+7=0
- x2+y2-6x+5=0
- x2+y2-4x+3=0
- x2+y2-10x+9=0
Answer
(2) x2+y2-6x+5=0
Solution
Shortest distance from (a,0) is along the normal at T(t2,2t): a=2+t2, PT2=4+4t2=16 t2=3, a=5.
[figure]
Focus S(1,0) and (5,0) lie on the axis, and the centre is on the axis they are the ends of a diameter.
Circle: (x-1)(x-5)+y2=0 x2+y2-6x+5=0
Answer: (2)