Assume e-4/5 = 2/5.If x, y satisfy, y = exand the minimum value of (x2+ y2) is expressed in the form of m/n then (2m - n)/5equals(where m & n are coprime natural numbers)Correct answer is '0.6'. Can you explain this answer?

Question

Answer

OP² = x² + y²
y = e^x, y' = e^x,

Answer

To solve this problem, we can start by solving the equation e-4/5 = 2/5 for e.

e-4/5 = 2/5

Multiply both sides by 5 to eliminate the fractions:

5(e-4/5) = 5(2/5)

e-4 = 2

Add 4 to both sides:

e = 2 + 4

e = 6

Now that we have the value of e, we can find the minimum value of (x^2 + y^2) using the given equation y = ex.

Substituting e = 6 into the equation:

y = 6x

Now we need to find the minimum value of (x^2 + y^2). We can rewrite this as:

(x^2 + (6x)^2)

Simplifying:

(x^2 + 36x^2)

37x^2

To find the minimum value of this expression, we can note that it is a quadratic equation in terms of x. The minimum value of a quadratic equation occurs at the vertex, which is given by x = -b/2a. In this case, a = 37 and b = 0, so the x-coordinate of the vertex is:

x = -0/(2*37)

x = 0

Substituting x = 0 into the expression:

37(0)^2 = 0

Therefore, the minimum value of (x^2 + y^2) is 0.

To express this in the form of m/n, we have:

(2m - n)/5 = (2(0) - 0)/5 = 0/5 = 0

So, (2m - n)/5 = 0.

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