S.O.S. Mathematics CyberBoard

Find the area of the triangle formed by the line with the eq. y=4 and the tangent lines to the parabola in the points

We know one of the sides of the triangle is the line y=4. We need to find the other two sides. Both sides are equal to the tangent of the equation y=x^2. The tangent of a point on a line is the derivative of the line. The derivative of x^2 is 2x, so at any point on x^2 the tangent line will be 2x. Using the two points (1,1) and (-2,4) we find the two tangent lines are y=2x and y=-4x. Those are the three sides of our triangle! Now just find the points where these 3 lines meet, find their lengths and then find the area!

Hello, PiDeltaPhi!

We have: y = x^2 and y' = 2x.

At (1,1), slope = 2. The equation of the tangent is y = 2x - 1
It intersects y = 4 at (5/2,4).

At (-2,4), slope = -4. The equation of the tangent is y = -4x - 4
It intersects y = 4 at (-2,4) and intersects the other tangent at (-1/2,-2).

Lucky for us, the triangle has a horizontal base: 5/2 - (-2) = 9/2
and a vertical height: 4 - (-2) = 6

The area of the triangle is: (1/2)(9/2)(6) = 27/2

Jzipp,

Soroban's solution is perfect.

While your idea is correct, some mistakes are found in your description.


This is incorrect. It is the gradient of the tangent at a point on the curve that is given by the derivative of y=x^2.


No.
At the point (1,1), the gradient of the tangent =2(1)=2,
hence the equation of the tangent is given by
y-1 = 2(x-1)
i.e. y=2x-1.

Similarly, at (-2,4), the equation of the tangent can be shown to be
y=-4x-4

As Homer would say - Doh!