Finding values with a given equation of a curve

Question

The curve y= (a²/(x+1))-b²x at the point (-2,2) has the gradient -25.

i) How do I find the values of a and b?

ii) How do I find the equation of the tangent line of the given curve at point x=1?

iii) What is the equation of the normal line at x=1?

Answer 1

y = a² / (x+1) - b²x

Since (-2,2) is on the curve, we have -a² + 2b² = 2.

dy/dx = -a² / (x+1)² - b²

Since dy/dx = -25 at the point (-2,2), -a² - b² = = -25

We have the system of equations:

-a² + 2b² = 2

-a² - b² = -25

Subtract the equations to get 3b² = 27. So, b² = 9 and a² = 16. Therefore, a = ±4 and b = ±3.

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When x = 1, y = 8 - 9 = -1

dy/dx = -4 - 9 = -13

Equation of tangent line at (1, -1): y + 1 = -13(x - 1). So, y = -13x + 12.

Equation of normal line at (1, - 1): y + 1 = (1/13)(x - 1). So, y = (1/13)x - 14/13.

Answer 2

Given that x= -2 and y = 2

Plug these values into the equation...

You will get:

-a^2 + 2b^2 = 2

Now, differentiate the equation with respect to x:

dy/dx = d/dx (a^2 / (x+1)) - d/dx (b^2 (x))

You will get:

dy/dx = -a^2 / (x+1) - b^2

At x= -2, dy/dx = -25

Plug these values into the derivative, and you will get the following equation:

-a^2 -b^2 = -25

You will have 2 equations and 2 unknowns:

-a^2 + 2b^2 = 2

-a^2 -b^2 = -25

Using substitution to solve for a and b, we get:

a^2 = 16

b^2 = 9

To find the point on the curve at x=1, plug this value into the equation of the curve...

y = 16 / 2 -9

y = -1

Now you have the point on the curve: (1,-1)

Plug x=1 into the equation of the derivative to find the slope...

dy/dx = -13

Use the point (1, -1 ) and the slope dy/dx = -13 to find the equation of the line...

y = -13x + 12

To find the equation of the normal line, we need to find a line that is perpendicular to y = -13x + 12

The slope of this perpendicular line should have the negative reciprocal slope of the tangent line, so

m= 1/13

We can use the slope of the tangent line and point on the curve (1, -1) to find the normal line:

y = (1/13)x - 14/13)