Sketch the curve $\displaystyle C$ with equation $\displaystyle y^2+4x^2=4+2y$(i have no problem with this part)
If the line $\displaystyle y=2x+a$ touches the curve $\displaystyle C$ at only one point, show that $\displaystyle a^2-2a-9=0$
Sketch the curve $\displaystyle C$ with equation $\displaystyle y^2+4x^2=4+2y$(i have no problem with this part)
If the line $\displaystyle y=2x+a$ touches the curve $\displaystyle C$ at only one point, show that $\displaystyle a^2-2a-9=0$
Substitute $\displaystyle \displaystyle y = 2x + a$ into the equation for $\displaystyle \displaystyle C$ and simplify. This will give you a Quadratic Equation.
There will only be one intersection when the discriminant of this quadratic $\displaystyle \displaystyle = 0$.
1. Calculate the point of intersection. (Replace the y in the 1st equation by the term of y of the 2nd equation:
$\displaystyle (2x-a)^2+4x^2=4+2(2x-a)~\implies~8x^2 + 4x(1 - a) + a^2 - 2a -4 = 0$
2. Solve for x: (Use the quadratic formula)
$\displaystyle x = \dfrac{-4(1-a)\pm\sqrt{(4(a-1))^2-4 \cdot 8 \cdot (a^2-2a-4)}}{2 \cdot 8}$
Expand the brackets in the root:
$\displaystyle x = \dfrac{-4(1-a)\pm\sqrt{-16(a^2-2a-9)}}{2 \cdot 8}$
3. YOU'll get only one point of intersection (the tangent point!) if the term in the root equals zero.
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.