Finding the Points of Intersection with the Coordinate Axes

Oct 13, 2025 by recoveryshake.com 129 views

Introduction

In this problem, we are given a parabola in the form of $$y=(k+x)(k-x)$$ and we need to find the value of $$k$$ such that the area of the triangle formed by joining the points where the parabola crosses the coordinate axes is 64 square units. To solve this problem, we will first find the points where the parabola crosses the coordinate axes, then we will use the formula for the area of a triangle to set up an equation and solve for $$k$$.

Finding the Points of Intersection with the Coordinate Axes

To find the points where the parabola crosses the coordinate axes, we need to find the values of $$x$$ and $$y$$ where the parabola intersects the $$x$$-axis and the $$y$$-axis.

When the parabola intersects the $$x$$-axis, the value of $$y$$ is 0. So, we can set up the equation:

0 = (k+x)(k-x)

Expanding the equation, we get:

0 = k^2 - x^2

Simplifying the equation, we get:

x^2 = k^2

Taking the square root of both sides, we get:

x = \pm k

So, the parabola intersects the $$x$$-axis at the points $$(k, 0)$$ and $$(-k, 0)$$.

When the parabola intersects the $$y$$-axis, the value of $$x$$ is 0. So, we can set up the equation:

y = (k+0)(k-0)

Simplifying the equation, we get:

y = k^2

So, the parabola intersects the $$y$$-axis at the point $$(0, k^2)$$.

Finding the Area of the Triangle

The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

In this case, the base of the triangle is the distance between the points $$(k, 0)$$ and $$(-k, 0)$$, which is $$2k$$. The height of the triangle is the distance between the point $$(0, k^2)$$ and the $$x$$-axis, which is $$k^2$$.

So, the area of the triangle is:

A = \frac{1}{2}(2k)(k^2)

Simplifying the equation, we get:

A = k^3

Setting Up an Equation and Solving for $$k$$

We are given that the area of the triangle is 64 square units. So, we can set up the equation:

k^3 = 64

Taking the cube root of both sides, we get:

k = \sqrt[3]{64}

Simplifying the equation, we get:

k = 4

So, the value of $$k$$ is 4.

Conclusion

In this problem, we found the points where the parabola $$y=(k+x)(k-x)$$ crosses the coordinate axes, then we used the formula for the area of a triangle to set up an equation and solve for $$k$$. We found that the value of $$k$$ is 4.

The Formula for the Area of a Triangle

The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

The Points of Intersection with the Coordinate Axes

The parabola intersects the $$x$$-axis at the points $$(k, 0)$$ and $$(-k, 0)$$.

The parabola intersects the $$y$$-axis at the point $$(0, k^2)$$.

The Value of $$k$$

The value of $$k$$ is 4.

The Area of the Triangle

The area of the triangle is 64 square units.

The Formula for the Area of a Triangle

The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

The Points of Intersection with the Coordinate Axes

The parabola intersects the $$x$$-axis at the points $$(k, 0)$$ and $$(-k, 0)$$.

The parabola intersects the $$y$$-axis at the point $$(0, k^2)$$.

The Value of $$k$$

The value of $$k$$ is 4.

The Area of the Triangle

The area of the triangle is 64 square units.

The Formula for the Area of a Triangle

The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

The Points of Intersection with the Coordinate Axes

The parabola intersects the $$x$$-axis at the points $$(k, 0)$$ and $$(-k, 0)$$.

The parabola intersects the $$y$$-axis at the point $$(0, k^2)$$.

The Value of $$k$$

The value of $$k$$ is 4.

The Area of the Triangle

The area of the triangle is 64 square units.

Q: What is the formula for the area of a triangle?

A: The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

Q: How do we find the points where the parabola crosses the coordinate axes?

A: To find the points where the parabola crosses the coordinate axes, we need to find the values of $$x$$ and $$y$$ where the parabola intersects the $$x$$-axis and the $$y$$-axis.

When the parabola intersects the $$x$$-axis, the value of $$y$$ is 0. So, we can set up the equation:

0 = (k+x)(k-x)

Expanding the equation, we get:

0 = k^2 - x^2

Simplifying the equation, we get:

x^2 = k^2

Taking the square root of both sides, we get:

x = \pm k

So, the parabola intersects the $$x$$-axis at the points $$(k, 0)$$ and $$(-k, 0)$$.

When the parabola intersects the $$y$$-axis, the value of $$x$$ is 0. So, we can set up the equation:

y = (k+0)(k-0)

Simplifying the equation, we get:

y = k^2

So, the parabola intersects the $$y$$-axis at the point $$(0, k^2)$$.

Q: How do we find the area of the triangle?

A: The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

So, the area of the triangle is:

A = \frac{1}{2}(2k)(k^2)

Simplifying the equation, we get:

A = k^3

Q: How do we set up an equation and solve for $$k$$?

A: We are given that the area of the triangle is 64 square units. So, we can set up the equation:

k^3 = 64

Taking the cube root of both sides, we get:

k = \sqrt[3]{64}

Simplifying the equation, we get:

k = 4

So, the value of $$k$$ is 4.

Q: What is the value of $$k$$?

A: The value of $$k$$ is 4.

Q: What is the area of the triangle?

A: The area of the triangle is 64 square units.

Q: What is the formula for the area of a triangle?

A: The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

Q: How do we find the points where the parabola crosses the coordinate axes?

A: To find the points where the parabola crosses the coordinate axes, we need to find the values of $$x$$ and $$y$$ where the parabola intersects the $$x$$-axis and the $$y$$-axis.

When the parabola intersects the $$x$$-axis, the value of $$y$$ is 0. So, we can set up the equation:

0 = (k+x)(k-x)

Expanding the equation, we get:

0 = k^2 - x^2

Simplifying the equation, we get:

x^2 = k^2

Taking the square root of both sides, we get:

x = \pm k

So, the parabola intersects the $$x$$-axis at the points $$(k, 0)$$ and $$(-k, 0)$$.

When the parabola intersects the $$y$$-axis, the value of $$x$$ is 0. So, we can set up the equation:

y = (k+0)(k-0)

Simplifying the equation, we get:

y = k^2

So, the parabola intersects the $$y$$-axis at the point $$(0, k^2)$$.

Q: How do we find the area of the triangle?

A: The area of a triangle is given by the formula:

A = \frac{1}{2}bh

where $$b$$ is the base of the triangle and $$h$$ is the height of the triangle.

So, the area of the triangle is:

A = \frac{1}{2}(2k)(k^2)

Simplifying the equation, we get:

A = k^3

Q: How do we set up an equation and solve for $$k$$?

A: We are given that the area of the triangle is 64 square units. So, we can set up the equation:

k^3 = 64

Taking the cube root of both sides, we get:

k = \sqrt[3]{64}

Simplifying the equation, we get:

k = 4

So, the value of $$k$$ is 4.