Question-18
The line \(\displaystyle y=x\) is a tangent to the parabola \(\displaystyle y=x^{2} +C\). Find \(\displaystyle C\).
Equate the slopes to begin with.
\(0.25\)
Let \(\displaystyle ( x_{0} ,x_{0})\) be the point of tangency. Since the slope of the tangent is \(\displaystyle 1\), we have\(\displaystyle 2x_{0} =1\Longrightarrow x_{0} =\frac{1}{2}\). Using the equation of the parabola, we get:
\[ \begin{equation*} \frac{1}{2} =\frac{1}{4} +C\Longrightarrow C=\frac{1}{4} \end{equation*} \]
Since the curves are simple enough, let us visualize this:
Thanks to Vivek, Sherry and Aniruddha for this solution:
The line \(\displaystyle y=x\) can intersect the parabola \(\displaystyle y=x^{2} +C\) at at most two points. Since \(\displaystyle y=x\) is a tangent to \(\displaystyle y=x^{2} +C\), there will just be one point of intersection. This means that the following quadratic equation has exactly one solution:
\[ \begin{equation*} x=x^{2} +C\Longrightarrow x^{2} -x+C=0 \end{equation*} \]
This happens when the discriminant of the quadratic expression is zero:
\[ \begin{equation*} 1-4C=0\Longrightarrow C=\frac{1}{4} \end{equation*} \]