Question
Find the shortest distance between the circle x^{2} + y^{2} – 24y + 128 = 0 and the parabola y^{2} = 4x.

None of these



easy
Solution
We know that shortest distance between two curve occurs along the common normal. Any normal to parabola will be of from
y + tx – 2t – t^{3} = 3. If it is common normal, of circle and parabola, it will passes through the center of the circle.
12 = 2t + t^{3}
(t – 2)(t^{2} + 2t + 6) = 0
t = 2
So point on parabola is (4, 4) distance of this point to center of a circle will be , hence shortest distance is
SIMILAR QUESTIONS
Show that the normal at a point (at^{2}, 2at) on the parabola y^{2} = 2ax cuts the curve again at the point whose parameter .
Find the locus of a pint P which moves such that two of the three normal’s drawn from it to the parabola y^{2} = 4ax are mutually perpendicular.
If normal at the point (at^{2}, 2at) in the parabola y^{2} = 4axintersects the parabola again at the (am^{2}, 2am), then find the minimum value of m^{2}.
The equation of circle touching the parabola y^{2} = 4x at the point (1, –2) and passing through origin is
The vertex of a parabola is the point (a, b) and latusrectum is of length l. If the axis of the parabola is along the positive direction of yaxis. Then its equation is
Slope of common tangent to parabolas y^{2} = 4x and x^{2} = 8y is
If a focal chord with positive slope of the parabola y^{2} = 16xtouches the circle x^{2} + y^{2} – 12x + 34 = 0, then m is
If 2y = x + 24 is a tangent to parabola y^{2} = 24x, then its distance from parallel normal is
PQ is a focal chord of the parabola y^{2} = 4ax, O is the origin. Find the coordinates of the centroid, G, of triangle OPQ and hence find the locus of G as PQ varies.
The equation of the directrix of the parabola y^{2} + 4y + 4x + 2 = 0 is