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# Solution for Find a Point on the Parabola Y = (X − 3)2, Where the Tangent is Parallel to the Chord Joining (3, 0) and (4, 1) ? - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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:

$f\left( x \right) = \left( x - 3 \right)^2 = x^2 - 6x + 9$

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 $\left( 4, 1 \right)$.
Is there an error in this question or solution?

#### APPEARS IN

Solution for 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) ? concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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