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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 19

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Answer: x y=2

Given: Let P(x, y) be the point of contact of tangent with curvey=f(x) which intercepts on \mathrm{x} \text {-axis }=2 x

To find: We have to show that the equation of curve which pass through \left ( 1,2 \right )

Hint: Use equation of tangent at point P\left ( x,y \right ) is  Y-y=\frac{d y}{d x}(X-x)

Solution: Equation of tangent at P\left ( x,y \right ) is  Y-y=\frac{d y}{d x}(X-x)

Put Y=0

        \begin{aligned} &=>-y=\frac{d y}{d x}(X-x) \\\\ &=>X-x=-y \frac{d x}{d y} \\\\ &=>X=x-y \frac{d x}{d y} \end{aligned}

Co-ordinate of  B\left(x-y \frac{d x}{d y}, 0\right)

Given( intercept on x-axis) = 2x

        \begin{aligned} &=>x-y \frac{d x}{d y}=2 x \\\\ &=>-y \frac{d x}{d y}=2 x-x \\\\ &=>-y \frac{d x}{d y}=x \end{aligned}

        =>-\frac{d x}{x}=\frac{d y}{y}        [separating variables]

On integrating on both sides, we get

        \begin{aligned} &=>-\int \frac{d x}{x}=\int \frac{d y}{y} \\\\ &=>-\log x=\log y+C \ldots(i) \end{aligned}

It is passing through \left ( 1,2 \right )

        \begin{aligned} &=>-\log 1=\log 2+C \\\\ &=>C=-\log 2 \end{aligned}

Put C=-\log 2 in equation (i) we get

        \begin{aligned} &=>-\log x=\log y-\log 2 \\\\ &=>\log x^{-1}=\log \left(\frac{y}{2}\right) \quad\quad\quad\left[-\log x=\log x^{-1}\right] \end{aligned}

        \begin{aligned} &=>\log \left(\frac{1}{x}\right)=\log \left(\frac{y}{2}\right) \\\\ &=>\frac{1}{x}=\frac{y}{2} \end{aligned}

Hence, xy=2 is required for equation of the curve.


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