SATHEE: Application of Derivatives 4 Question 48

Application of Derivatives 4 Question 48

####50. The circle $x^{2} + y^{2} = 1$ cuts the $X$ -axis at $P$ and $Q$ . Another circle with centre at $Q$ and variable radius intersects the first circle at $R$ above the $X$ -axis and the line segment $P Q$ at $S$ . Find the maximum area of the $△ Q S R$ .

(1994, 5M)

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Answer:

Correct Answer: 50. $(0, 2)$

Solution:

Since $x^{2} + y^{2} = 1$ a circle $S_{1}$ has centre $(0, 0)$ and cuts $X$ -axis at $P (- 1, 0)$ and $Q (1, 0)$ . Now, suppose the circle $S_{2}$ , with centre at $Q (1, 0)$ has radius $r$ . Since, the circle has to meet the first circle, $0 < r < 2$ .

Again, equation of the circle with centre at $Q (1, 0)$ and radius $r$ is

$(x - 1)^{2} + y^{2} = r^{2}$

To find the coordinates of point $R$ , we have to solve it with

$x^{2} + y^{2} = 1$

On subtracting Eq. (ii) from Eq. (i), we get

$\begin{aligned} (x - 1)^{2} - x^{2} & = r^{2} - 1 \\ \Rightarrow & x^{2} + 1 - 2 x - x^{2} & = r^{2} - 1 \\ \Rightarrow & 1 - 2 x & = r^{2} - 1 \\ ∴ & x & = \frac{2 - r^{2}}{2} \end{aligned}$

On putting the value of $x$ in Eq. (i), we get

$\begin{aligned} \frac{2 - r^{2}}{2} + y^{2} = 1 \end{aligned}$

$\begin{aligned} = 1 - \frac{r^{4} - 4 r^{2} + 4}{4} \\ = \frac{4 - r^{4} + 4 r^{2} - 4}{4} \\ = \frac{4 r^{2} - r^{4}}{4} \\ = \frac{r^{2} (4 - r^{2})}{4} \\ \Rightarrow y = \frac{r \sqrt{4 - r^{2}}}{2} \end{aligned}$

Again, we know that, coordinates of $S$ are $(1 - r, 0)$ , therefore

$S Q = 1 - (1 - r) = r$

Let $A$ denotes the area of $△ Q S R$ , then

$\begin{aligned} A & = \frac{1}{2} r r \frac{\sqrt{4 - r^{2}}}{2} \\ = \frac{1}{4} r^{2} \sqrt{4 - r^{2}} \\ \Rightarrow & A^{2} = \frac{1}{16} r^{4} (4 - r^{2}) \\ Let & f (r) = r^{4} (4 - r^{2}) = 4 r^{4} - r^{6} \\ \Rightarrow & f^{'} (r) = 16 r^{3} - 6 r^{5} = 2 r^{3} (8 - 3 r^{2}) \end{aligned}$

For maxima and minima, put $f^{'} (r) = 0$

$\begin{aligned} \Rightarrow & 2 r^{3} (8 - 3 r^{2}) & = 0 \\ \Rightarrow & r & = 0, 8 - 3 r^{2} = 0 \\ \Rightarrow & r & = 0, 3 r^{2} = 8 \\ \Rightarrow & r & = 0, r^{2} = 8 / 3 \\ \Rightarrow & r = 0, r & = \frac{2 \sqrt{2}}{\sqrt{3}} \end{aligned}$

$[∵ 0 < r < 2, so r = 2 \sqrt{2} / \sqrt{3}]$

Again, $f^{''} (r) = 48 r^{2} - 30 r^{4}$

$\begin{aligned} \Rightarrow f^{''} \frac{2 \sqrt{2}}{\sqrt{3}} & = 48 \frac{4 \times 2}{3} - 30 \frac{4 \times 2}{3}^{2} \\ = 16 \times 8 - \frac{10 \times 64}{3} = 128 - \frac{640}{3} = - \frac{256}{2} < 0 \end{aligned}$

Therefore, $f (r)$ is maximum when, $r = \frac{2 \sqrt{2}}{\sqrt{3}}$

Hence, maximum value of $A$

$\begin{aligned} = \frac{1}{4} {\frac{2 \sqrt{2}}{\sqrt{3}}}_{2}^{4 - {\frac{2 \sqrt{2}}{\sqrt{3}}}^{2}} = \frac{1}{4} \frac{8}{3} \cdot \sqrt{4 - \frac{8}{3}} \\ = \frac{2}{3} \cdot \frac{\sqrt{12 - 8}}{\sqrt{3}} = \frac{2 \cdot 2}{3 \sqrt{3}} = \frac{4}{3 \sqrt{3}} = \frac{4 \sqrt{3}}{9} \end{aligned}$

is smallest at $x = 1$ .

So, $f (x)$ is decreasing on $[0, 1]$ and increasing on $[1, 3]$ . Here, $f (1) = - 1$ is the smallest value at $x = 1$ .

$∴$ Its smallest value occur as

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (- x^{3}) + \frac{(b^{3} - b^{2} + b - 1)}{b^{2} + 3 b + 2}$

In order this value is not less than -1 , we must have

$\begin{aligned} \frac{b^{3} - b^{2} + b - 1}{b^{2} + 3 b + 2} \geq 0 \\ \Rightarrow \frac{(b^{2} + 1) (b - 1)}{(b + 1) (b + 2)} \geq 0 \end{aligned}$

Application of Derivatives 4 Question 48

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Application of Derivatives 4 Question 48

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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