A normal is drawn to `(x)/(9)+(y)/(4)=2` .at a point (3,2) .The perpendicular distance of normal from origin is :

The pedal points of a perpendicular drawn from origin on the line 3x+4y-5=0, is

Normal at (2, 2) to curve x^2 + 2xy – 3y^2 = 0 is L. Then perpendicular distance from origin to line L is

Find the equation of a curve passing through the point (1.1) if the perpendicular distance of the origin from the normal at any point P(x,y) of the curve is equal to the distance of P from the x-axis.

A point P(x,y) nores on the curve x^((2)/(3))+y^((2)/(3))=a^((2)/(3)),a>0 for each position (x,y) of p,perpendiculars are drawn from origin upon the tangent and normal at P,the length (absolute valve) of them being p_(1)(x) and P_(2)(x) brespectively,then

A curve passing through the point (1,2) has the property that the perpendicular distance of the origin from normal at any point P of the curve is equal to the distance of P from the x-axis is a circle with radius.

If normals are drawn to the ellipse x^(2)+2y^(2)=2 from the point (2,3). then the co-normal points lie on the curve

A normal is drawn to the ellipse (x^(2))/(9)+y^(2)=1 at the point (3cos theta, sin theta) where 0lt theta lt(pi)/(2) . If N is the foot of the perpendicular from the origin O to the normal such that ON = 2, then theta is equal to

The locus of the point from which perpendicular tangent and normals can be drawn to a circle is

A normal is drawn to the parabola y^(2)=9x at the point P(4, 6). A circle is described on SP as diameter, where S is the focus. The length of the intercept made by the circle on the normal at point P is :

Three normals drawn to the parabola y^(2) = 4x from the point (c, 0) are real and diferent if