Line touches curve at only one point

**Punch** · February 25th 2011, 04:01 AM

Sketch the curve $C$ with equation $y^2+4x^2=4+2y$ (i have no problem with this part)

If the line $y=2x+a$ touches the curve $C$ at only one point, show that $a^2-2a-9=0$

**Prove It** · February 25th 2011, 04:51 AM

Substitute $\displaystyle y = 2x + a$ into the equation for $\displaystyle C$ and simplify. This will give you a Quadratic Equation.

There will only be one intersection when the discriminant of this quadratic $\displaystyle = 0$ .

**earboth** · February 25th 2011, 05:01 AM

Originally Posted by Punch

Sketch the curve $C$ with equation $y^2+4x^2=4+2y$ (i have no problem with this part)

If the line $y=2x+a$ touches the curve $C$ at only one point, show that $a^2-2a-9=0$

1. Calculate the point of intersection. (Replace the y in the 1st equation by the term of y of the 2nd equation:

$(2x-a)^2+4x^2=4+2(2x-a)~\implies~8x^2 + 4x(1 - a) + a^2 - 2a -4 = 0$

2. Solve for x: (Use the quadratic formula)

$x = \dfrac{-4(1-a)\pm\sqrt{(4(a-1))^2-4 \cdot 8 \cdot (a^2-2a-4)}}{2 \cdot 8}$

Expand the brackets in the root:

$x = \dfrac{-4(1-a)\pm\sqrt{-16(a^2-2a-9)}}{2 \cdot 8}$

3. YOU'll get only one point of intersection (the tangent point!) if the term in the root equals zero.

**Punch** · February 25th 2011, 11:26 PM

Originally Posted by earboth

1. Calculate the point of intersection. (Replace the y in the 1st equation by the term of y of the 2nd equation:

$(2x-a)^2+4x^2=4+2(2x-a)~\implies~8x^2 + 4x(1 - a) + a^2 - 2a -4 = 0$

2. Solve for x: (Use the quadratic formula)

$x = \dfrac{-4(1-a)\pm\sqrt{(4(a-1))^2-4 \cdot 8 \cdot (a^2-2a-4)}}{2 \cdot 8}$

Expand the brackets in the root:

$x = \dfrac{-4(1-a)\pm\sqrt{-16(a^2-2a-9)}}{2 \cdot 8}$

3. YOU'll get only one point of intersection (the tangent point!) if the term in the root equals zero.

thanks! but how do i show that $a^2-2a-9=0$ ?

**Prove It** · February 25th 2011, 11:32 PM

As was pointed out twice, if there's only one point of intersection, then the discriminant of your resulting quadratic has to be 0.

The discriminant is all the stuff under the square root.

**HallsofIvy** · February 26th 2011, 02:27 AM

If the discriminant of the quadratic equation is negative, there will be no (real number) solution- the line will NOT intersect the circle. If the discriminant of the quadratic equation is positive, there will be two (real number) solutions- the line will intersect the circle twice. If the discriminant of the quadratic equation is zero, there will be exactly one (real number) solution- the line will intersect the circle exactly once. And that was what you were told was to happen.

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