Question

Find a point on parabola $f\left(x\right)=(x-3{)}^2$, where tangent is parallel to the chord joining the points (3,0) and (4,1)

Answer 1

$f\left(x\right)={(x-3)}^2.....\left(1\right)\left(\text{Given}\right)$

The given points are $(3,0)$ and $(4,1)$.

Therefore,

$f\left(x\right)={(x-3)}^2;x\in [3,4]$

Since $f\left(x\right)$ is a polynomial function, $f\left(x\right)$ is continuous in $[3,4]$.

$f\text{'}\left(x\right)=2(x-3)$ which exists in $(3,4)$

Therefore it is also differentiable in $(3,4)$.

Hence both the conditions of Lagrange's Mean value theorem is satisfied.

Therefore,

$f\left(3\right)={(3-3)}^2=0$

$f\left(4\right))={(4-3)}^2=1$

$\therefore f\text{'}\left(c\right)=2(c-3)$

Therefore,

$2c-6=\frac{1-0}{4-3}$

$\Rightarrow 2c-6=1$

$\Rightarrow c=\frac{7}{2}$

$\frac{7}{2}\in (3,4)$

Since $c$ is the x-coordinate of that point at which the tangent is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

Therefore Substituting $x=\frac{7}{2}$ in ${eq}^n\left(1\right)$, we have

$f\left(x\right)={(\frac{7}{2}-3)}^2=\frac{1}{4}$

Thus the required point is $(\frac{7}{2},\frac{1}{4})$.

Hence the correct answer is $(\frac{7}{2},\frac{1}{4})$.