At what point in the first quadrant on the parabola y=4-x^2 does the tangent line, together with the coordinate axis, determine a triangle of minimum area?
Let the point of contact of the tangent be P(a,b)
then b = 4 - a^2
we know dy/dx = -2x, which is the slope of the tangent
so at P the slope is -2a
equation of the tangent is
y = -2ax + k
but (a,b) lies on it
b = -2a(a) + k
k = b + 2a^2 = 4-a^2 + 2a^2
k = 4 + a^2
tangent equation : y = -2ax + 4+a^2
the base of the triangle is the x-intercept
the height of the triangle is the y-intercept
x-intercept: let y = 0
0 = -2ax + 4 + a^2
x = (4 + a^2)/(2a)
y-intercept: let x = 0
y = 4 + a^2
area = (1/2)(4+a^2)/(2a) (4+a^2) = (4+a^2)^2/(4a)
your turn!
Take the derivative of this using the quotient rule
Set that equal to zero and solve for a,
Sub back into the top to get b for the point P(a,b)
Sub into “area” to get the minimum area.
I got a = 2/√3
To find the point on the parabola y = 4 - x^2 where the tangent line, together with the coordinate axis, forms a triangle of minimum area, we need to find the point where the derivative of y is equal to zero.
Step 1: Find the derivative of y.
The derivative of y = 4 - x^2 can be obtained by applying the power rule of differentiation. Differentiating each term separately, we get:
dy/dx = d(4)/dx - d(x^2)/dx
= 0 - 2x
= -2x
Step 2: Set the derivative equal to zero and solve for x.
Setting -2x = 0, we find that x = 0.
Step 3: Substitute the value of x into the original equation to find the corresponding y-coordinate.
Substituting x = 0 into y = 4 - x^2, we get y = 4.
Therefore, the point on the parabola y = 4 - x^2 where the tangent line, together with the coordinate axis, determines a triangle of minimum area is (0, 4).
To find the point on the parabola y = 4 - x^2 where the tangent line, together with the coordinate axis, determines a triangle of minimum area, we need to use calculus.
Step 1: Determine the equation for the tangent line
Let's start by finding the derivative of the equation y = 4 - x^2 to get dy/dx.
dy/dx = -2x
The derivative gives us the slope of the tangent line at any point on the parabola.
Step 2: Set up the equation of the tangent line
We can use the point-slope form of a line to determine the equation of the tangent line at a general point (x, y) on the parabola. We will use the slope we found in Step 1 and the point (x, y) to get the equation of the tangent line.
y - (4 - x^2) = -2x * (x - x)
Simplifying the equation, we get:
y = 2x^2 + 4
Step 3: Calculate the points where the tangent line crosses the coordinate axis
To determine the points where the tangent line crosses the x and y-axis, we set y = 0 for the x-axis and x = 0 for the y-axis.
For the x-axis, we set y = 0 and solve for x:
0 = 2x^2 + 4
2x^2 = -4
x^2 = -2
This equation has no real solutions, which means the tangent line doesn't cross the x-axis.
For the y-axis, we set x = 0 and solve for y:
y = 2(0)^2 + 4
y = 4
So, the line crosses the y-axis at the point (0, 4).
Step 4: Find the coordinates of the point where the tangent line intersects the parabola
Now we need to find the x-coordinate of the point where the tangent line intersects the parabola.
Since the equation of the tangent line is y = 2x^2 + 4, we can set this equal to the equation of the parabola, 4 - x^2.
2x^2 + 4 = 4 - x^2
3x^2 = 0
x^2 = 0
This means x = 0, which gives us the coordinate (0, 4) as the point where the tangent line intersects the parabola.
Step 5: Calculate the area of the triangle
To find the area of the triangle, we can use the formula for the area of a triangle, which is 1/2 * base * height.
In this case, the base is the x-coordinate of the point where the tangent line intersects the parabola, which is 0. The height is the y-coordinate of the point where the tangent line intersects the y-axis, which is 4.
Area = 1/2 * 0 * 4
Area = 0
Therefore, the triangle has a minimum area of 0.
To summarize, at the point (0, 4) on the parabola y = 4 - x^2, the tangent line, together with the coordinate axis, determines a triangle of minimum area, which is 0.