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Find a Point on the Parabola Y = (X − 4)2, Where the Tangent is Parallel to the Chord Joining (4, 0) and (5, 1) ? - Mathematics

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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) ?

Sum
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Solution

​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).
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Chapter 15: Mean Value Theorems - Exercise 15.2 [Page 18]

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RD Sharma Mathematics [English] Class 12
Chapter 15 Mean Value Theorems
Exercise 15.2 | Q 5 | Page 18

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