The Condition for Tangency

Oct 13, 2025 by magicalceylontours.com 94 views

Introduction

In this article, we will explore the problem of finding the positive value of $m$ for the line $y+9=mx$ that is tangent to the curve $4y-x^2+4=0$ . To solve this problem, we will use the concept of tangency and the properties of quadratic equations.

Understanding the Problem

The given curve is a quadratic equation in the form of $4y-x^2+4=0$ . We are asked to find the positive value of $m$ for the line $y+9=mx$ that is tangent to this curve. This means that the line intersects the curve at exactly one point, and we need to find the slope of this line.

The Condition for Tangency

For the line $y+9=mx$ to be tangent to the curve $4y-x^2+4=0$ , it must intersect the curve at exactly one point. This means that the quadratic equation formed by substituting the equation of the line into the equation of the curve must have exactly one solution.

Substituting the Equation of the Line into the Equation of the Curve

We can substitute the equation of the line $y+9=mx$ into the equation of the curve $4y-x^2+4=0$ to get:

4(mx-9)-x^2+4=0

Simplifying this equation, we get:

4mx-36-x^2+4=0

Rearranging the terms, we get:

x^2-4mx+32=0

The Quadratic Formula

The quadratic formula states that the solutions to the quadratic equation $ax^2+bx+c=0$ are given by:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

In this case, we have:

a=1, b=-4m, c=32

Substituting these values into the quadratic formula, we get:

x=\frac{4m\pm\sqrt{(-4m)^2-4(1)(32)}}{2(1)}

Simplifying this expression, we get:

x=\frac{4m\pm\sqrt{16m^2-128}}{2}

The Condition for One Solution

For the line $y+9=mx$ to be tangent to the curve $4y-x^2+4=0$ , the quadratic equation $x^2-4mx+32=0$ must have exactly one solution. This means that the discriminant $b^2-4ac$ must be equal to zero.

Setting the Discriminant Equal to Zero

We can set the discriminant $b^2-4ac$ equal to zero and solve for $m$ :

(-4m)^2-4(1)(32)=0

Simplifying this expression, we get:

16m^2-128=0

Solving for $m$

We can solve for $m$ by dividing both sides of the equation by 16:

m^2-8=0

Adding 8 to both sides of the equation, we get:

m^2=8

Taking the square root of both sides of the equation, we get:

m=\pm\sqrt{8}

Simplifying the Square Root

We can simplify the square root by factoring out a perfect square:

m=\pm\sqrt{4\cdot2}

Simplifying this expression, we get:

m=\pm2\sqrt{2}

The Positive Value of $m$

We are asked to find the positive value of $m$ . Therefore, we take the positive square root:

m=2\sqrt{2}

Conclusion

In this article, we have found the positive value of $m$ for the line $y+9=mx$ that is tangent to the curve $4y-x^2+4=0$ . The positive value of $m$ is $2\sqrt{2}$ .

Final Answer

The final answer is: $\boxed{2\sqrt{2}}$

Introduction

In our previous article, we explored the problem of finding the positive value of $m$ for the line $y+9=mx$ that is tangent to the curve $4y-x^2+4=0$ . We used the concept of tangency and the properties of quadratic equations to find the positive value of $m$ . In this article, we will answer some frequently asked questions related to this problem.

Q: What is the condition for the line $y+9=mx$ to be tangent to the curve $4y-x^2+4=0$ ?

A: The line $y+9=mx$ must intersect the curve $4y-x^2+4=0$ at exactly one point. This means that the quadratic equation formed by substituting the equation of the line into the equation of the curve must have exactly one solution.

Q: How do we find the equation of the quadratic that is formed by substituting the equation of the line into the equation of the curve?

A: We can substitute the equation of the line $y+9=mx$ into the equation of the curve $4y-x^2+4=0$ to get:

4(mx-9)-x^2+4=0

Simplifying this equation, we get:

4mx-36-x^2+4=0

Rearranging the terms, we get:

x^2-4mx+32=0

Q: What is the discriminant of the quadratic equation $x^2-4mx+32=0$ ?

A: The discriminant of the quadratic equation $x^2-4mx+32=0$ is given by:

b^2-4ac=(-4m)^2-4(1)(32)=16m^2-128

Q: What is the condition for the quadratic equation $x^2-4mx+32=0$ to have exactly one solution?

A: The quadratic equation $x^2-4mx+32=0$ must have exactly one solution. This means that the discriminant $b^2-4ac$ must be equal to zero.

Q: How do we find the value of $m$ that satisfies the condition for the quadratic equation $x^2-4mx+32=0$ to have exactly one solution?

A: We can set the discriminant $b^2-4ac$ equal to zero and solve for $m$ :

16m^2-128=0

Simplifying this expression, we get:

m^2-8=0

Adding 8 to both sides of the equation, we get:

m^2=8

Taking the square root of both sides of the equation, we get:

m=\pm\sqrt{8}

Q: What is the positive value of $m$ that satisfies the condition for the quadratic equation $x^2-4mx+32=0$ to have exactly one solution?

A: The positive value of $m$ that satisfies the condition for the quadratic equation $x^2-4mx+32=0$ to have exactly one solution is:

m=2\sqrt{2}

Q: What is the final answer to the problem of finding the positive value of $m$ for the line $y+9=mx$ that is tangent to the curve $4y-x^2+4=0$ ?

A: The final answer to the problem of finding the positive value of $m$ for the line $y+9=mx$ that is tangent to the curve $4y-x^2+4=0$ is:

m=2\sqrt{2}

Conclusion

In this article, we have answered some frequently asked questions related to the problem of finding the positive value of $m$ for the line $y+9=mx$ that is tangent to the curve $4y-x^2+4=0$ . We hope that this article has been helpful in clarifying any doubts that you may have had about this problem.

Final Answer

The final answer is: $\boxed{2\sqrt{2}}$