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 \[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 3 \right)}{4 - 3}\].

Now, 

\[f\left( x \right) = x^2 - 6x + 9\]\[\Rightarrow\] \[f'\left( x \right) = 2x - 6\],\[f\left( 3 \right) = 0, f\left( 4 \right) = 1\]\[\therefore\] \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 3 \right)}{4 - 3}\]

\[\Rightarrow\] \[2x - 6 = \frac{1 - 0}{4 - 3} \Rightarrow 2x = 7 \Rightarrow x = \frac{7}{2}\]

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

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 \[\left( 4, 1 \right)\].