Find a point on the curve `y=x^3+1` where the tangent is parallel to the chord joining `(1,\ 2)` and `(3,\ 28)` .

Answer

Points on the curve are `(frac{sqrt{13}}{sqrt{3}}, frac{13 sqrt{13}+3 sqrt{3}}{3 sqrt{3}})`

Explanation

It is given that tangent is parallel to chord
`\therefore` slope of tangent `=` slope of chord
Let `m_{1}=` slope of tangent `\quad` and `m_{2}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=` slope of line joining 2 points
`y=x^{3}+1 \quad m_{2}=\frac{26}{2}=13`
$$ \frac{d y}{d x}=3 x^{2}=m_{1} $$
because `m_{1}=m_{2}`
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