#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 33 maths textbook solution

Answer: $x^{2}-y^{2}=-a^{2}$

Given: The product of the slope and the ordinate is equal to the abscissa i.e. y(slope of tangent)

= x

To find: The equation of the curves that pass through the point $\left ( 0,a \right )$.

Hint: Use  $y \times \text { slope of tangent }=x$ to solve.

Solution: we have y(slope of tangent) = x

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

Integrating on both sides we get

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

The curve passes through $\left ( 0,a \right )$

\begin{aligned} &=\frac{a^{2}}{2}=\frac{0^{2}}{2}+C \\\\ &=\frac{a^{2}}{2}=C \\\\ &=C=\frac{a^{2}}{2} \end{aligned}

Substituting in equation (i) we get

$=\frac{y^{2}}{2}=\frac{x^{2}}{2}+\frac{a^{2}}{2}$

Multiply 2 on both sides

\begin{aligned} &=y^{2}=x^{2}+a^{2} \\\\ &=x^{2}-y^{2}=a^{2} \end{aligned}

Hence, required equation of curve is found.

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