The subnormal to the curve xy = c2 at any point varies directly asa) ...
Statement: The subnormal to the curve xy = c^2 at any point varies directly as the cube of the ordinate(y^3).
Explanation:
To understand the solution, we need to know what subnormal is, and how it is related to the given curve.
Subnormal: The line segment drawn from the point of contact on the curve to the x-axis, perpendicular to the tangent at that point is called the subnormal.
In the given curve, let P(x, y) be any point on the curve. Then, the equation of the curve can be written as y = c^2/x.
Now, let us find the slope of the tangent to the curve at point P.
Differentiating y = c^2/x with respect to x, we get
(dy/dx) * x = -c^2/x^2
=> dy/dx = -c^2/x^3
Therefore, the slope of the tangent at point P is -c^2/x^3.
Now, let us find the equation of the tangent at point P.
Equation of the tangent at point P can be written as
y - y1 = m(x - x1)
=> y - c^2/x = -c^2/x^3(x - x)
=> y = -c^2/x^2 + 2c^2/x
Now, let us find the subnormal at point P.
The subnormal is perpendicular to the tangent and passes through the point P.
Let Q be the point of contact of the subnormal with the curve. Then, the equation of the subnormal can be written as
(y - y1) = -1/m(x - x1)
=> y - c^2/x = x^3/c^2
=> y = x^3/c^2 + c^2/x
Now, we need to show that the subnormal varies directly as the cube of the ordinate.
Let y = kx^3/c^2 + c^2/x, where k is a constant.
Differentiating y with respect to x, we get
(dy/dx) = 3kx^2/c^2 - c^2/x^2
Therefore, the slope of the subnormal is (dy/dx) = 3kx^2/c^2 - c^2/x^2.
Now, we know that the subnormal is perpendicular to the tangent at point P. Therefore, the product of their slopes is -1.
=> (-c^2/x^3) * (3kx^2/c^2 - c^2/x^2) = -1
=> 3kx^4 - c^4 = -x^5
=> k = -1/3c^4
Therefore, y = (-1/3c^4) * x^3/c^2 + c^2/x
=> y = -x^3/(3c^6) + c^2/x
Hence, we can see that the subnormal varies directly as the cube of the ordinate.
Therefore, the correct answer is option 'A'.