Question

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 1

It is given that tangent is parallel to chord

$\therefore$ slope of tangent = slope of chord

Let $m_1=$ slope of tangent and $m_2=\frac{y_2-y_1}{x_2-x_1}=$ slope of line joining 2 points

$y=x^3+1$ $m_2=\frac{26}{2}=13$

$\frac{dy}{dx}=3x^2=m_1$

$\because m_1=m_2$

$3x^2=13$

$x=\sqrt{\frac{13}{3}}$

$\therefore y={\left(\sqrt{\frac{13}{3}}\right)}^3+1$

$y=\frac{13}{3}\frac{\sqrt{13}}{\sqrt{3}}+1$

$\therefore$ Points on the curve are $(\frac{\sqrt{13}}{\sqrt{3}},\frac{13\sqrt{13}+3\sqrt{3}}{3\sqrt{3}})$