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

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}

$=\frac{16}{2}=9-\frac{9}{2}+C$

\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.