SATHEE: Application of Derivatives 4 Question 9

Application of Derivatives 4 Question 9

####10. Let $A (4, - 4)$ and $B (9, 6)$ be points on the parabola, $y^{2} = 4 x$ . Let $C$ be chosen on the arc $A O B$ of the parabola, where $O$ is the origin, such that the area of $△ A C B$ is maximum. Then, the area (in sq. units) of $△ A C B$ , is

(2019 Main, 9 Jan II)

(a) $31 \frac{1}{4}$

(b) 32

(d) $30 \frac{1}{2}$

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Answer:

Correct Answer: 10. (a)

Solution:

According to given information, we have the following figure.

For $y^{2} = 4 a x$ , parametric coordinates of a point is $(a t^{2}$ , $2 a t)$ .

$∴$ For $y^{2} = 4 x$ , let coordinates of $C$ be $(t^{2}, 2 t)$ .

Then, area of $△ A B C = \frac{1}{2} | | \begin{array}{ccc} t^{2} & 2 t & 1 \\ 9 & 6 & 1 \\ 4 & - 4 & 1 \end{array} ∣$

$= \frac{1}{2} | t^{2} (6 - (- 4)) - 2 t (9 - 4) + 1 (- 36 - 24) |$

$= \frac{1}{2} | 10 t^{2} - 10 t - 60 | = \frac{10}{2} | t^{2} - t - 6 | = 5 | t^{2} - t - 6 |$

Let, $A (t) = 5 | t^{2} - t - 6 |$

Clearly, $A (4, - 4) \equiv A (t_{1}^{2}, 2 t_{1}) \Rightarrow 2 t_{1} = - 4$

$\Rightarrow t_{1} = - 2$

and $B (9, 6) \equiv B (t_{2}^{2}, 2 t_{2}) \Rightarrow 2 t_{2} = 6 \Rightarrow t_{2} = 3$

Since, $C$ is on the arc $A O B$ , the parameter ’ $t$ ’ for point $C \in (- 2, 3)$ .

Let $f (t) = t^{2} - t - 6 \Rightarrow f^{'} (t) = 2 t - 1$

Now, $f^{'} (t) = 0 \Rightarrow t = \frac{1}{2}$

Thus, for $A (t)$ , critical point is at $t = \frac{1}{2}$

Now, $A \frac{1}{2} = 5 | {\frac{1}{2}}^{2} - \frac{1}{2} - 6 | = \frac{125}{4} = 31 \frac{1}{4}$ [Using Eq. (i)]

Application of Derivatives 4 Question 9

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Application of Derivatives 4 Question 9

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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