A normal is drawn at a point `P(x,y)` of a curve. It meets the x-axis at `Q` such that `PQ` is of constant length `k`. Answer the question:If the curve passes through the point `(0,k)`, then its equation is (A) `x^2-y^2=k^2` (B) `x^2+y^2=k^2` (C) `x^2-y^2=2k^2` (D) `x^2+y^2=2k^2`

A normal is drawn at a point P(x,y) of a curve It meets the x-axis at Q If PQ is of constant length k such a curve passing through (0,k) is

A normal is drawn at a point P(x,y) of a curve. It meets the x-axis at Q such that PQ is of constant length k . Answer the question:The differential equation describing such a curve is (A) y dy/dx=+-sqrt(k^2-x^2) (B) x dy/dx=+-sqrt(k^2-x^2) (C) y dy/dx=+-sqrt(k^2-y^2) (D) x dy/dx=+-sqrt(k^2-y^2)

A normal is drawn at a point P(x,y) of a curve.It meets the x-axis at Q. If PQ has constant length k, then show that the differential equation describing such curves is y(dy)/(dx)=+-sqrt(k^(2)-y^(2)). Find the equation of such a curve passing through (0,k).

If y=2x+k is a tangent to the curve x^(2)=4y , then k is equal to

If the tangent at the point (p,q) to the curve x^(3)+y^(2)=k meets the curve again at the point (a,b), then

If 3x+4y-k is the equation of the normal to the curve x^(2)+2x-3y+3=0 at the point (1,2) ,then the value of k=

The equation of the curve,slope of whose tangent at any point (h,k) is (2k)/(h) and which passes through the point (1,1) is (i)x^(2)=y( ii) y^(2)=x( iii) x^(2)=2y( iv )y^(2)=2x

If a circle of constant radius 3k passes through the origin O and meets the coordinate axes at AandB,then the locus of the centroud of triangle OAB is (a)x^(2)+y^(2)=(2k)^(2)(b)x^(2)+y^(2)=(3k)^(2)(c)x^(2)+y^(2)=(4k)^(2)(d)x^(2)+y^(2)=(6k)^(2)