Q. 243.5( 2 Votes )

# 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.

Answer :

⇒

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 – 2k^{2} = cx – ch – 2hx + 2h^{2}

⇒ x(c – 2h) – 2ky + 2k^{2} –hc + 2h^{2} = 0

⇒ x(c – 2h) – 2ky + 2(k^{2} –2h) – hc = 0

⇒ x(c – 2h) – 2ky + 2(ch) – hc = 0 ( h^{2} + k^{2} = ch as (h,k) on th curve)

⇒ x(c – 2h) – 2ky + hc = 0

Now, Length of perpendicular as tangent from origin is

Hence, x^{2} + y^{2} = cx is the required curve.

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