# Find the point of intersection of the tangents to the curve y = x^2 at the points (-1/2, 1/4) and (1, 1). y= x^2

The slope of tangent line of the curve y is the first derivative at the point ( -1/2, 1/4).

==> y= 2x.

==> m = 2*-1/2 = -1

==> (y-1/4) = m ( x+1/2)

==> y-1/4 = -1(x+1/2)

==> y= -x - 1/2 + 1/4

==> y= -x...

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y= x^2

The slope of tangent line of the curve y is the first derivative at the point ( -1/2, 1/4).

==> y= 2x.

==> m = 2*-1/2 = -1

==> (y-1/4) = m ( x+1/2)

==> y-1/4 = -1(x+1/2)

==> y= -x - 1/2 + 1/4

==> y= -x -1/4

==> 4y + 4x + 1 = 0

Now we will find the tangent line at the point (1,1).

==< Then, the slope is m = 2*1 = 2

==> (y-1) = 2(x-1)

==> y= 2x - 2 + 1

==> y= 2x -1

Now we will determine the intersection points between both lines.

==> y= 2x - 1

==> y = -x -1/4

==> 2x -1 = -x - 1/4

==> 3x = -1/4 + 1

==> 3x = 3/4

==> x = 1/4

==> y= -1/2.

The, the intersection point is the point ( 1/4, -1/2),

Approved by eNotes Editorial Team The first derivative of a curve gives the slope of the tangent to the curve at any point.

We are given the curve y = x^2

Now y’ = 2x.

At the point (-1/2, 1/4), y’ = 2*(-1/2) = -1

The equation of the tangent is y – (-1/2) = -1*(x – 1/4)

=> y + 1/2 = 1/4 – x

=> x + y + 1/4 = 0

=> 4x + 4y +1 = 0

At the point (1, 1), y’ = 2* 1 = 2

The equation of the tangent is y -1 = 2*(x – 1)

=> y – 1 = 2x – 2

=> 2x – y – 1 = 0

To determine the point of intersection of 4x + 4y +1 = 0 and 2x – y – 1 = 0

2x – y – 1 = 0

=> y = 2x – 1

substitute this in 4x + 4y +1 = 0

=> 4x + 4(2x – 1) +1 =0

=> 4x + 8x – 4 + 1 = 0

=> 12x – 3 = 0

=> x = 3/12

=> x = 1/4

As y = 2x – 1

=> y = 2*(1/4) – 1

=> y = 1/2 – 1

=> y = -1/2

Therefore the point of intersection of the required tangents is (1/4, -1/2).

Approved by eNotes Editorial Team