If two normals to a parabola `y^2 = 4ax` intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are:

A normal chord of the parabola y^2=4ax subtends a right angle at the vertex if its slope is

If a, b, c are in AP then ax + by + c = 0 will always pass through a fixed point whose coordinates are

If the parabola y^2 = ax passes through (3,2) then the focus is

If the two tangents drawn from a point P to the parabola y^(2) = 4x are at right angles then the locus of P is

Show that all chords of the curve 3x^2-y^2-2x+4y=0, which subtend a right angle at the origin, pass through a fixed point. Find the coordinates of the point.

The ellipse 4x^2+9y^2=36 and the hyperbola a^2x^2-y^2=4 intersect at right angles. Then the equation of the circle through the points of intersection of two conics is

Circle are drawn on the chords of the rectangular hyperbola x y=4 parallel to the line y=x as diameters. All such circles pass through two fixed points whose coordinates are (2,2) (b) (2,-2) (c) (-2,2) (d) (-2,-2)

If a normal to a parabola y^2 =4ax makes an angle phi with its axis, then it will cut the curve again at an angle

If the parabola y=(a-b)x^2+(b-c)x+(c-a) touches x- axis then the line ax+by+c=0 passes through a fixed point