The slope of tangent at any point (x, y) on a curve y = y ( x ) is x 2 + y 2 2 x y , x 0 . If y ( 2 ) = 0 , then a value of y ( 8 ) is [2023]

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Solution

Q.
The slope of tangent at any point (x, y) on a curve $y = y (x) is \frac{x^{2} + y^{2}}{2 x y}, x > 0 .$ If $y (2) = 0$ , then a value of $y (8)$ is [2023]

1 $- 2 \sqrt{3}$

2 $2 \sqrt{3}$

3 $4 \sqrt{3}$

4 $- 4 \sqrt{2}$

Ans.
(3)

$We have, \frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$

Putting $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ , we get

$v + x \frac{d v}{d x} = \frac{1}{2} (\frac{1 + v^{2}}{v})$

$\Rightarrow x \frac{d v}{d x} = \frac{1}{2} (\frac{1}{v} - v) = \frac{1}{2} (\frac{1 - v^{2}}{v})$ $∴$ $\int \frac{2 v}{1 - v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \log c - \log$ $| 1 - v^{2} |$ $= \log$ $| x |$

$c = x (1 - \frac{y^{2}}{x^{2}}) = \frac{x^{2} - y^{2}}{x}$

At $y (2) = 0, c = 2$

$∴$ $\frac{x^{2} - y^{2}}{x} = 2$

Now, at $x = 8,$ $2 = \frac{8^{2} - y^{2}}{8} \Rightarrow y^{2} = 64 - 16 = 48 \Rightarrow y = 4 \sqrt{3}$

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