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Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 26

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Answer: x^{2}+y^{2}-6 x-7=0

Given: Equation of the normal at point \left ( x,y \right ) on the curve (Y-y)=-\frac{d x}{d y}(X-x)

To find: we have to find the equation of curve which passes through the point \left ( 3,0 \right ) if the curve contains the point \left ( 3,4 \right )

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

Solution: Equation of normal on point \left ( x,y \right ) on the curve is

        =Y-y=-\frac{d x}{d y}(X-x)

It is passing through \left ( 3,0 \right )

        \begin{aligned} &=0-y=-\frac{d x}{d y}(3-x) \\\\ &=y=\frac{d x}{d y}(3-x) \\\\ &=y d y=(3-x) d x \end{aligned}

Integrating on both sides

        \begin{aligned} &=\int y d y=\int(3-x) d x \\\\ &=\int y d y=\int 3 d x-\int x d x \\\\ &=\frac{y^{2}}{2}=3 x-\frac{x^{2}}{2}+C \ldots(i) \end{aligned}

It passes through \left ( 3,4 \right )

        \begin{aligned} &=\frac{4^{2}}{2}=3 \times 3-\frac{3^{2}}{2}+C \\\\ \end{aligned}


        \begin{aligned} &=\frac{16}{2}=\frac{18-9}{2}+C \\\\ &=\frac{16}{2}=\frac{9}{2}+C \end{aligned}

        \begin{aligned} &=C=\frac{16-9}{2} \\\\ &=C=\frac{7}{2} \end{aligned}    

Substituting C=\frac{7}{2}  in equation (i)

         =\frac{y^{2}}{2}=3 x-\frac{x^{2}}{2}+\frac{7}{2}   

Multiplying by 2

        =y^{2}=6 x-x^{2}+7

Hence, the required equation is found.



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