Find The Shortest Distance Between The Circle x2 + y2 – 24y + 128 = 0 And The Parabola y2 = 4x.

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Find the shortest distance between the circle x2 + y2 – 24y + 128 = 0 and the parabola y2 = 4x.


Correct option is

We know that shortest distance between two curve occurs along the common normal. Any normal to parabola will be of from

y + tx – 2t – t3 = 3. If it is common normal, of circle and parabola, it will passes through the center of the circle.

 12 = 2t + t3

 (t – 2)(t2 + 2t + 6) = 0

   = 2

So point on parabola is (4, 4) distance of this point to center of a circle will be , hence shortest distance is 



Show that the normal at a point (at2, 2at) on the parabola y2 = 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 y2 = 4ax are mutually perpendicular.


If normal at the point (at2, 2at) in the parabola y2 = 4axintersects the parabola again at the (am2, 2am), then find the minimum value of m2.


The equation of circle touching the parabola y2 = 4x at the point  (1, –2) and passing through origin is


The vertex of a parabola is the point (a, b) and latus-rectum is of length l. If the axis of the parabola is along the positive direction of y-axis. Then its equation is


Slope of common tangent to parabolas y2 = 4x and x2 = 8y is


If a focal chord with positive slope of the parabola y2 = 16xtouches the circle x2 + y2 – 12+ 34 = 0, then m is


If 2y = x + 24 is a tangent to parabola y2 = 24x, then its distance from parallel normal is


PQ is a focal chord of the parabola y2 = 4axO 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 y2 + 4y + 4x + 2 = 0 is