Looks easy but still Dangerous.

Calculus Level 5

A curve passing through the origin, is such that the middle point of the segment of normal between point of contact and $x$ -axis lies on the parabola $2y^2 = x$ , then the equation the curve is $a_{1}y^{a_{2}} = a_{3}x + a_{4} - a_{5}e^{a_{6}x}$ . Find the value of $a_{1} +a_{2} + a_{3} +a_{4} + a_{5} + a_{6}$ .

Given that $a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}$ are integers which may or may not be equal with $a_{1}$ > 0 and their HCF is 1.

Find more trouble here .

The answer is 9.

1 solution

Tom Engelsman
Jul 31, 2016

Let y = f(x) be the curve in question with f(0) = 0 as a boundary condition. The normal at any point P(k, f(k)) can be written as:

y - f(k) = (-1/f ' (k)) (x - k) => y = (-1/f ' (k)) x + [k/f ' (k) + f(k)] (i)

which has its x-intercept (i.e. y = 0) at the point Q(k + f(k)*f ' (k), 0) and midpoint M equal to:

M = ( (k + k + f(k) f ' (k))/2, (f(k) + 0)/2 ) = (k + f(k) f ' (k)/2, f(k)/2) (ii).

If M always lies on the parabola 2*y^2 = x, then we obtain:

2 [f(k)/2]^2 = k + f(k) f ' (k)/2 => f(k)^2 / 2 = k + f(k) f ' (k)/2 => f(k)^2 = 2k + f(k) f ' (k) => y^2 = 2x + yy' (iii).

Now if one makes the substitution: u = (y^2)/2 and u' = yy', then (iii) is transformed into the first-order ODE:

2u = 2x + u' => u' - 2u = -2x (iv).

Taking exp(-2x) for an integrating factor, we now integrate (iv) into:

u exp(-2x) = (x + 1/2) exp(-2x) + C => u = x + 1/2 + C exp(2x) => (y^2)/2 = x + 1/2 + C exp(2x) => y^2 = 2x + 1 + 2C*exp(2x) (v).

where C is a real constant of integration. Finally, plugging in the boundary condition y(0) = 0 produces C = -1/2, which leaves us with:

y^2 = 2x + 1 - exp(2x) (vi)

with a1 = a4 = a5 = 1 and a2 = a3 = a6 = 2 => a1 + a2 + a3 + a4 + a5 + a6 = 9.

Looks easy but still Dangerous.

The answer is 9.

1 solution

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