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

justaguide | College Teacher | (Level 2) Distinguished Educator

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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).

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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),

neela | High School Teacher | (Level 3) Valedictorian

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To find the intersection of tangents to y = x^2.

The equation of tangent to a curve at (x1,y1) is given by:

y-y1 = (dy/dx)(x-x1).

So dy/dx  = (x^2)' = 2x.

At x= (-1/2), dy/dx = 2(-1/2) =  -1.

So the tangent at (-1/2, 1/4) is given by:

y-1/4 = (-1)(x+1/2). Or

y-1/4 = -x-1/2.

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

y = -x-1/4...................(1).

Similarly at x = 1, dy/dx = 2x = 2*1 = 2. So the tangent at (1,1) is given by:

y-1 = 2(x-1)

=>y = 2x-2+1 = 2x-1. Or

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

y = -x-1/4.........(1)

From (1) and (2) we get:

2x-1 = -x-1/4

=>2x+x = -1/4 +1 = 3/4.

=>3x= 3/4.

=> x = (3/4)/3 = 1/4.

Put x= 1/4 in eq (1) and we get  y = -x-1/4 = -1/4-1/4 = -1/2.

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