## Question

If normal at the point (*at*^{2}, 2*at*) in the parabola *y*^{2} = 4*ax*intersects the parabola again at the (*am*^{2}, 2*am*), then find the minimum value of *m*^{2}.

### Solution

8

Since normal at the point ‘*t*’ intersects again at the point ‘*m*’

We have

*t*^{2} +* tm* + 2 = 0

Since ‘*t*’ is real, *D* ≥ 0

So, least value of *m*^{2} is 8.

#### SIMILAR QUESTIONS

The vertex of the parabola *y*^{2} = 8*x* is at the center of a circle and the parabola cuts the circle at the ends of its latus rectum. Then the equation of the circle is

Find the equation of the parabola whose focus is (1, 1) and the directrix is *x + y *+ 1 = 0.

If the line 2*x* + 3*y* = 1 touch the parabola *y*^{2} = 4*ax* at the point*P*. Find the focal distance of the point *P*.

Find the angle between the tangents of the parabola *y*^{2} = 8*x*, which are drawn from the point (2, 5).

Find the locus of middle point of chord *y*^{2} = 4*ax* drawn through vertex.

Find the locus of the mid-point of the chords of the parabola *y*^{2} = 4*ax* which subtend a right angle at the vertex of the parabola.

Show that the normal at a point (*at*^{2}, 2*at*) on the parabola *y*^{2} = 2*ax* cuts the curve again at the point whose parameter .

Show that the normal at a point (*at*^{2}, 2*at*) on the parabola *y*^{2} = 2*ax* 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} = 4*ax* are mutually perpendicular.

The equation of circle touching the parabola *y*^{2} = 4*x* at the point (1, –2) and passing through origin is