Let \(P\) be the point on the curve \(y=ax^2+bx+c\) (where \(a\) is non-zero) at which the gradient is \(m\). Show that the equation of the tangent at \(P\) is \[y-mx=c-\frac{(m-b)^2}{4a}.\]

What is the gradient of the quadratic anywhere on the curve? What is it at \(P\)? Therefore, what are the coordinates of \(P\)?

Show that the curves \(y=a_1x^2+b_1x+c_1\) and \(y=a_2x^2+b_2x+c_2\) (where \(a_1\) and \(a_2\) are non-zero) have a common tangent with gradient \(m\) if and only if \[(a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+4a_1a_2(c_2-c_1)+a_2b_1^2-a_1b_2^2=0.\]

The rest of this question requires you to think what statements imply others (and whether it works in reverse). Curves have a common tangent if and only if the equations of the tangents are the same if and only if gradients and intercepts are the same.

It may be helpful for clarity to think about the two directions separately: show that if they have a common tangent then the equation holds, and that if the equation holds then they have a common tangent.

Show that, in the case \(a_1\neq a_2\), the two curves have exactly one common tangent if and only if they touch each other.

This part requires arguments in both the forward \(\Longrightarrow\) direction and in the backward \(\Longleftarrow\) direction to get the full ‘if and only if’.

Forward direction (left to right, \(\Longrightarrow\)): Curves have *one* common tangent, so we want only one solution to the quadratic \[(a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+4a_1a_2(c_2-c_1)+a_2b_1^2-a_1b_2^2=0.\] What is the discriminant?

If the two quadratics touch each other, we want only one solution to \[a_1x^2+b_1x+c_1=a_2x^2+b_2x+c_2.\]

In the case \(a_1=a_2\), find a necessary and sufficient condition for the two curves to have exactly one common tangent.

What is meant by a necessary and sufficient condition?

What is a necessary and sufficient condition for a linear equation of the form \(dx+e=f\) to have a solution for \(x\)?