If length of the perpendicular from the origin upon the tangent drawn to the curve `x^2 - xy + y^2 + alpha (x-2)=4` at `(2, 2)` is equal to 2 then `alpha` equals

The length of the perpendicular from the origin, on the normal to the curve, x^2 + 3xy - 10 y^2 = 0 at the point (2,1) is :

The length of the perpendicular from the origin, on the normal to the curve, x^2 + 3xy - 10 y^2 = 0 at the point (2,1) is :

The length of the perpendicular from the origin,on the normal to the curve, x^(2)+2xy-3y^(2)=0 at the point (2,2) is

If p_(1) and p_(2) be the lengths of the perpendiculars from the origin upon the tangent and normal respectively to the curve x^((2)/(3)) +y^((2)/(3)) = a^((2)/(3)) at the point (x_(1), y_(1)) , then-

Product of perpendiculars drawn from the foci upon any tangent to the ellipse 3x^(2)+4y^(2)=12 is

The feet of the perpendicular drawn from focus upon any tangent to the parabola,y=x^(2)-2x-3 lies on

If p and q are the lengths of the perpendiculars from the origin to the straight lines x "sec" alpha + y " cosec" alpha = a " and " x "cos" alpha-y " sin" alpha = a "cos" 2alpha, " then prove that 4p^(2) + q^(2) = a^(2).

A curve is such that the length of perpendicular from origin on the tangent at any point P of the curve is equal to the abscissa of P . Prove that the differential equation of the curve is y^2-2xy dy/dx-x^2=0 and hence find the curve.

The point of contact of the tangents drawn from origin to the curve y=x^2+3x+4 is

Let OM be the perpendicular from origin upon tangent at any point P on the curve (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 then the length of the normal at P varies as