# A 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 P. Prove that the differential equation of the curve is and hence find the curve.

It is a homogenous equation,

Putting y = kx

So,

Putting the value of k

Differentiating with respect to x,

Let (h,k) be the point where tangent passes through origin and the length is equal to h. So, equation of tangent at (h,k) is

2ky – 2k2 = cx – ch – 2hx + 2h2

x(c – 2h) – 2ky + 2k2 –hc + 2h2 = 0

x(c – 2h) – 2ky + 2(k2 –2h) – hc = 0

x(c – 2h) – 2ky + 2(ch) – hc = 0 ( h2 + k2 = ch as (h,k) on th curve)

x(c – 2h) – 2ky + hc = 0

Now, Length of perpendicular as tangent from origin is

Hence, x2 + y2 = cx is the required curve.

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