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, then show that the differential equation describing such curves is, y 
= ± and the equation of such a curve passing through (0, k) -
Text Solution
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Let y = ƒ(x) be the curve such that the normal at P (x, y) to this curve meets x-axis at Q. Then,
PQ = length of the normal at P.
= y 
But PQ = k
∴ y 
= k
⇒ y 2 + y 2 
= k 2
Or y 
= ± 
, Integrating both sides, we get 
,
– 
= ± x + c, since it passes through (0, k) → c = 0. ∴ – 
= ± x
or k 2 – y 2 = x 2
⇒ x 2 + y 2 = k 2 , is required equation of the curve
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