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

If p is the length of the perpendicular from a focus upon the tangent at any point P of the the ellipse x^2/a^2+y^2/b^2=1 and r is the distance of P from the focus , then (2a)/r-(b^2)/(p^2) is equal to

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

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

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.

If p is the length of the perpendicular from a focus upon the tangent at any point P of the the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and rs the distance of P from the foicus,then (2a)/(r)-(b^(2))/(p^(2)) is equal to

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

Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse x^2/a^2+y^2/b^2=1 is b^2

Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse x^2/a^2+y^2/b^2=1 is b^2

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 :