23. [-/2 Points] DETAILS MY NOTES SCALCET9 3.1.081. For what values of \( a \) and \( b \) is the line \( 4 x+y=b \) tangent to the parabola \( y=a x^{2} \) when \( x=5 \) ? \( a=\square \) \( b= \)

1. First, write the line in slope-intercept form: \[ 4x + y = b \quad \Longrightarrow \quad y = -4x + b. \] 2. The parabola is given by: \[ y = ax^2. \] 3. Since the line is tangent to the parabola at \( x = 5 \), the two curves must intersect at \( x = 5 \). Therefore, the \( y \)-coordinates are the same at \( x = 5 \). For the parabola: \[ y = a(5)^2 = 25a. \] For the line: \[ y = -4(5) + b = -20 + b. \] Setting these equal gives: \[ 25a = -20 + b. \] 4. The slope of the line is \( -4 \). At the point of tangency, the derivative of the parabola must equal the slope of the line. Differentiate \( y = ax^2 \) with respect to \( x \): \[ y' = 2ax. \] At \( x = 5 \), the derivative becomes: \[ y'\big|_{x=5} = 10a. \] Setting the slope equal to the line's slope: \[ 10a = -4. \] 5. Solve for \( a \): \[ a = -\frac{4}{10} = -\frac{2}{5}. \] 6. Substitute \( a = -\frac{2}{5} \) into the equation from Step 3: \[ 25\left(-\frac{2}{5}\right) = -20 + b. \] Simplify: \[ -10 = -20 + b. \] Solve for \( b \): \[ b = -10 + 20 = 10. \] 7. Therefore, the values are: \[ a = -\frac{2}{5} \quad \text{and} \quad b = 10. \]