Find the equation of the tangent to the curve y=2x² which is parallel to the secant of the curve which passes through the points on the curve which have the coordinates x=−1 and x=2.
Solution
$ \displaystyle \begin{array}{l}\text{Curve : }y=2{{x}^{2}}\\\\\text{When }x=-1,y=2{{(-1)}^{2}}=2\\\\\text{When }x=2,y=2{{(2)}^{2}}=8\\\\\therefore \text{Gradient of secant = }\frac{{8-2}}{{2-(-1)}}=2\\\\\text{The gradient of tangent is }\frac{{dy}}{{dx}}.\\\\\therefore \ \frac{{dy}}{{dx}}=4x\\\\\text{By the problem, tangent }\parallel \ \text{secant}\text{.}\\\\\therefore \ \frac{{dy}}{{dx}}=2\Rightarrow 4x=2\Rightarrow x=\frac{1}{2}\\\\\text{When }x=\frac{1}{2},y=2{{(\frac{1}{2})}^{2}}=\frac{1}{2}\end{array}$
$\displaystyle \begin{array}{l}\therefore \text{The tangent line touches the curve at (}\frac{1}{2},\frac{1}{2}\text{)}\text{.}\\\\\therefore \text{Equation of tangent line to the curve at (}\frac{1}{2},\frac{1}{2}\text{)}\ \text{is}\\\\\ \ y-\ \frac{1}{2}=2(x-\frac{1}{2})\Rightarrow y=2x-\frac{1}{2}.\end{array}$
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