 # Find a point on the parabola `
Question:

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

Solution:

​Let:

$f(x)=(x-3)^{2}=x^{2}-6 x+9$

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

Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$

$\therefore a=3, b=4$

The polynomial function is everywhere continuous and differentiable.

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

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

Consequently, there exists $c \in(3,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(3)}{4-3}$.

Now,

$f(x)=x^{2}-6 x+9 \Rightarrow f^{\prime}(x)=2 x-6, f(3)=0, f(4)=1$

$\therefore f^{\prime}(x)=\frac{f(4)-f(3)}{4-3} \Rightarrow 2 x-6=\frac{1-0}{4-3} \Rightarrow 2 x=7 \Rightarrow x=\frac{7}{2}$

Thus, $c=\frac{7}{2} \in(3,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(3)}{4-3}$.

Clearly,

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

Thus, $(c, f(c))$, 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 $(3,0)$ and $(4,1)$.