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) ?

Answer 1

​Let: 

\[f\left( x \right) = \left( x - 4 \right)^2 = x^2 - 8x + 16\] 

The tangent to the curve is parallel to the chord joining the points \[\left( 4, 0 \right)\] and \[\left( 5, 1 \right)\] .

Assume that the chord joins the points 

\[\left( a, f\left( a \right) \right)\] and \[\left( b, f\left( b \right) \right)\] .

\[\therefore\] \[a = 4, b = 5\]

The polynomial function is everywhere continuous and differentiable.
So, \[x^2 - 8x + 16\] is continuous on \[\left[ 4, 5 \right]\] and differentiable on \[\left( 4, 5 \right)\] .

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists \[c \in \left( 4, 5 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 5 \right) - f\left( 4 \right)}{5 - 4}\] .

Now, 

\[f\left( x \right) = x^2 - 8x + 16 \Rightarrow\] \[f'\left( x \right) = 2x - 8\],\[f\left( 5 \right) = 1, f\left( 4 \right) = 0\]

\[\therefore\]  \[f'\left( x \right) = \frac{f\left( 5 \right) - f\left( 4 \right)}{5 - 4}\]\[\Rightarrow\] \[2x - 8 = \frac{1}{1} \Rightarrow 2x = 9 \Rightarrow x = \frac{9}{2}\]

Thus, \[c = \frac{9}{2} \in \left( 4, 5 \right)\] such that ​\[f'\left( c \right) = \frac{f\left( 5 \right) -f\left( 4 \right)}{5 - 4}\] .

Clearly,

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

Thus, \[\left( c, f\left( c \right) \right)\] 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).