 # Find a point on the parabola `
Question:

Find a point on the parabola $y=(x-4)^{2}$, where the tangent is parallel to the chord joining $(4,0)$ and $(5,1)$.

Solution:

Let:

$f(x)=(x-4)^{2}=x^{2}-8 x+16$

The tangent to the curve is parallel to the chord joining the points $(4,0)$ and $(5,1)$.

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

$\therefore a=4, b=5$

The polynomial function is everywhere continuous and differentiable.

So, $x^{2}-8 x+16$ is continuous on $[4,5]$ and differentiable on $(4,5)$.

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

Consequently, there exists $c \in(4,5)$ such that $f^{\prime}(c)=\frac{f(5)-f(4)}{5-4}$.

Now,

$f(x)=x^{2}-8 x+16 \Rightarrow f^{\prime}(x)=2 x-8, f(5)=1, f(4)=0$

$\therefore f^{\prime}(x)=\frac{f(5)-f(4)}{5-4} \Rightarrow 2 x-8=\frac{1}{1} \Rightarrow 2 x=9 \Rightarrow x=\frac{9}{2}$

Thus, $c=\frac{9}{2} \in(4,5)$ such that $f^{\prime}(c)=\frac{f(5)-f(4)}{5-4}$

Clearly,

$f(c)=\left(\frac{9}{2}-4\right)^{2}=\frac{1}{4}$

Thus, $(c, f(c))$, i.e. $\left(\frac{9}{2}, \frac{1}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $(4,0)$ and $(5,1)$