A curve y = f(x) passes through (1,1) and tangent at P(x,y) cuts the x-axis and y- axis at A and B, respectively, such that BP : AP = 3:1 ,a)equation of curve is xy’ - 3y = 0b)normal at (1,1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 8Correct answer is option 'C,D'. Can you explain this answer?

Question

A curve y = f(x) passes through (1,1) and tangent at P(x,y) cuts the x-axis and y- axis at A and B, respectively, such that BP : AP = 3:1 ,a)equation of curve is xy&rsquo; - 3y = 0b)normal at (1,1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 8Correct answer is option 'C,D'. Can you explain this answer?

Answer

To find the equation of the curve, we need to determine the slope of the tangent line at point P.

Let's assume the coordinates of point P are (x, y). Since the curve passes through (1, 1), we can find the slope of the tangent line at P using the formula:

slope = (y - 1) / (x - 1)

Since the line BP:AP = 3:1, the ratio of the distances from P to B and P to A is 3:1. Let's assume the distance from P to B is 3k and the distance from P to A is k.

We can find the coordinates of points A and B using the given information. The x-coordinate of point A is 0, and the y-coordinate of point B is 0. Therefore, we have the following equations:

y - k = f(x)
y - 3k = f'(x) * (x - 1)

To find the equation of the curve, we need to eliminate the variable y from these equations. Let's solve the second equation for y:

y = f'(x) * (x - 1) + 3k

Substitute this expression for y in the first equation:

f(x) = f'(x) * (x - 1) + 3k - k
f(x) = f'(x) * (x - 1) + 2k

Now, let's substitute the expression for f(x) in the equation xy = f(x):

xy = (f'(x) * (x - 1) + 2k) * x

Simplifying this equation:

xy = f'(x) * x^2 - f'(x) * x + 2k*x

This is the equation of the curve.

Answer

Tangent to the curve y = f(x) at (x,y) is

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