# Thread: Line touches curve at only one point

1. ## Line touches curve at only one point

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 $a^2-2a-9=0$

2. 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 $\displaystyle = 0$.

3. 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.

4. 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 $a^2-2a-9=0$?

5. 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.

6. 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.