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

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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...

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neela | Student

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).

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