SATHEE: Application of Derivatives 4 Question 47

Application of Derivatives 4 Question 47

####49. Let $(h, k)$ be a fixed point, where $h > 0, k > 0$ . A straight line passing through this point cuts the positive directions of the coordinate axes at the points $P$ and $Q$ . Find the minimum area of the $△ O P Q, O$ being the origin.

(1995, 5M)

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Answer:

Correct Answer: 49. $\frac{3 \sqrt{3}}{4} r^{2}$ sq. units

Solution:

Let equation of any line through the point $(h, k)$ is

$y - k = m (x - h)$

For this line to intersect the positive direction of two axes, $m = \tan θ < 0$ , since the angle in anti-clockwise direction from $X$ -axis becomes obtuse.

The line (i) meets $X$ -axis at $P h - \frac{k}{m}, 0$ and $Y$ -axis at $Q (0, k - m h)$ .

Let $A =$ area of $△ O P Q = \frac{1}{2} O P \cdot O Q$

$\begin{aligned} = \frac{1}{2} h - \frac{k}{m} (k - m h) \\ = \frac{1}{2} \frac{m h - k}{m} (k - m h) = - \frac{1}{2 m} (k - m h)^{2} \\ = - \frac{1}{2 \tan θ} (k - h \tan θ)^{2} [∵ m = \tan θ] \\ = - \frac{1}{2 \tan θ} (k^{2} + h^{2} \tan^{2} θ - 2 h k \tan θ) \\ = \frac{1}{2} (2 k h - k^{2} \cot θ - h^{2} \tan θ) \end{aligned}$

$\Rightarrow \frac{d A}{d θ} = \frac{1}{2} [- k^{2} (- {cosec}^{2} θ) - h^{2} \sec^{2} θ]$

$= \frac{1}{2} [k^{2} {cosec}^{2} θ - h^{2} \sec^{2} θ]$

To obtain minimum value of $A, \frac{d A}{d θ} = 0$

$\Rightarrow k^{2} {cosec}^{2} θ - h^{2} \sec^{2} θ = 0$

$\Rightarrow \frac{k^{2}}{\sin^{2} θ} = \frac{h^{2}}{\cos^{2} θ} \Rightarrow \frac{k^{2}}{h^{2}} = \tan^{2} θ$

$\Rightarrow \tan θ = \pm \frac{k}{h}$

$∵ \tan θ < 0, k > 0, h > 0$

[given]

Therefore, $\tan θ = - \frac{k}{h}$ (only possible value).

Now, $\frac{d^{2} A}{d θ^{2}} = \frac{1}{2} [- 2 k^{2} {cosec}^{2} θ \cot θ - 2 h^{2} \sec^{2} θ \tan θ]$

$= - [k^{2} (1 + \cot^{2} θ) \cot θ + h^{2} (1 + \tan^{2} θ) \tan θ]$

$\Rightarrow \frac{d^{2} A}{d θ^{2}}_{\tan θ = - k / h} = - k^{2} 1 + \frac{h^{2}}{k^{2}} \frac{- h}{k}$

$+ h^{2} 1 + \frac{k^{2}}{h^{2}} \frac{- k}{h}$

$= k^{2} \frac{k^{2} + h^{2}}{k^{2}} \frac{h}{k} + h^{2} \frac{h^{2} + k^{2}}{h^{2}} \frac{k}{h}$

$= \frac{(k^{2} + h^{2}) h}{k} + \frac{(h^{2} + k^{2}) (k)}{h}$

$= (k^{2} + h^{2}) \frac{h}{k} + \frac{k}{h} > 0$

$[∵ h, k > 0]$

Therefore, $A$ is least when $\tan θ = - k / h$ . Also, the least value of $A$ is

$\begin{aligned} A & = \frac{1}{2} 2 h k - k^{2} \frac{- h}{k} - h^{2} \frac{- k}{h} \\ = \frac{1}{2} [2 h k + k h + k h] = 2 h k \end{aligned}$

Application of Derivatives 4 Question 47

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Application of Derivatives 4 Question 47

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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