Question

Find a point on the parabola y = (x − 3)², where the tangent is parallel to the chord joining (3, 0) and (4, 1).

Answer 1

​Let:
$f\left(x\right)={\left(x-3\right)}^2=x^2-6x+9$

The tangent to the curve is parallel to the chord joining the points $\left(3,0\right)$ and $\left(4,1\right)$.

Assume that the chord joins the points $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right)$.

$\therefore$ $a=3,b=4$

The polynomial function is everywhere continuous and differentiable.

So, $f\left(x\right)=x^2-6x+9$ is continuous on $\left[3,4\right]$ and differentiable on $\left(3,4\right)$.

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c\in \left(3,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$.

Now,
$f\left(x\right)=x^2-6x+9$$\Rightarrow$$f\text{'}\left(x\right)=2x-6$, $f\left(3\right)=0,f\left(4\right)=1$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$$\Rightarrow$$2x-6=\frac{1-0}{4-3}\Rightarrow 2x=7\Rightarrow x=\frac{7}{2}$

Thus, $c=\frac{7}{2}\in \left(3,4\right)$ such that ​$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$.

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

Thus, $\left(c,f\left(c\right)\right)$, i.e. $\left(\frac{7}{2},\frac{1}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $\left(3,0\right)$ and $\left(4,1\right)$.