# Prove that the graph of the curve y=x^2 is tangent to the graph of the curve y=x+1-1/x.Write the equation of the tangent line in the intercepting point.

neela | Student

We shall see whether the curve  y = x^2 touches or intersect the curve  y= x+1-1/x.

At the point of inersection (or may be contact), the ordinates are equal.

So y = x^2 = x+1-1/x

x^3= x^2+x-1

x^3-x^2 -x+1 = 0

x^2(x-1)-1(x-1) = 0

(x-1)(x^2-1) = 0.

(x+1)(x-1)^2 = 0

So x =1 is a double point.

Let us find the slope dy/dx for y = x^2 at x= 1.

dy/ dx = 2x . So slope at x = 1  for y = x^2 is  m1 = {(2x) at x=1 } = 2.

Let us find the slope for y = x+1-1/x at x = 1.

dy/dx = (x+1-1/x)' = {(1+1/x^2 )at x= 1} = 1+1/1 = 2.

Thus  x=1 is point where the curves touch as it is double point and the slope  dy/dx = 2 for both curves.

giorgiana1976 | Student

To prove that the graphs of the given curves are tangent to each other, we'll solve the equation:

x^2 = x + 1 - 1/x

We'll multiply the term from the left side and the first 2 terms from the right side, by x;

x^3 = x^2 + x - 1

We'll move all terms to one side:

x^3 - x^2 - x + 1 = 0

We'll factorize by x^2 the first 2 terms:

x^2*(x-1) - (x-1) = 0

We'll factorize again by (x-1):

(x-1)*(x^2-1) = 0

We'll write the difference of squares (x^2 - 1) = (x-1)(x+1)

We'll re-write the product:

(x-1)*(x-1)*(x+1) = 0

We'll put each factor as zero:

x-1 = 0

x = 1

x+1 = 0

x = -1

Since y = x^2, we'll have;

For x = 1 => y = 1

For x = -1 => y = 1

So, the curves are tangent. The point where the 2 curves are tangent is (1 , 1).

Now, we'll write the equation of the line that passes through a point and it has a slope m.

The slope m = y'(1).

y = x^2 => y' = 2x

y'(1) = 2*1

y'(1) = 2

y'(1) = m

m = 2

The equation of the line is:

y - 1 = y'(1)* (x-1)

y - 1 = 2(x-1)

We'll remove the brackets:

y-1-2x+2=0

We'll combine like terms and the equation of the line, tangent to the curves, is:

y - 2x + 1 = 0