Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 93

Find the values of $a$ and $b$ such that the line $2x+y = b$ is tangent to the parabola $y = ax^2$ when $x=2$

If the line is tangent to the curve at some point. It means that the first derivative of the
curve is equal t the slope of the line and by inspection, the slope of the line is -2 using the
general equation of the line $y = mx + b $ where $m$ is the slope.

Taking the first derivative of the curve we get

$
\begin{equation}
\begin{aligned}
y &= ax^2\\
\\
y' &= 2ax && \text{;when } x = 2\\
\\
y' &= 2a(2)\\
\\
y' &= 4a && \text{;equating this to the slope of the line}\\
\\
-2 &= 4a\\
\\
a &= \frac{-1}{2}
\end{aligned}
\end{equation}
$


To find the value of $b$, we equate both equations since they have a point of intersection


$
\begin{equation}
\begin{aligned}
2x + y &= b \\
\\
y &= -2x + b \quad ; \text{when } x=2 \qquad y = -2(2)+b \qquad = -4 +b\\
\\
y &= ax^2 \qquad ; \text{when } x = 2 \qquad y = a(2)^2 \qquad =4a\\
\\
4a &= -4 + b \qquad \text{Substituting the value of } a\\
\\
4\left(\frac{-1}{2}\right) &= -4+b \qquad \text{Solve for } b\\
\\
b &= 2
\end{aligned}
\end{equation}
$


Therefore the required values are $\displaystyle a = -\frac{1}{2}$ and $b = 2$