#### please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1 sub question 5 maths textbook solution

$\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$

Hint - The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

m1=slope of first curve.   m2=slope of second curve.

Given –

$\begin{array}{r} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \ldots \ldots \text { (1) } \\ x^{2}+y^{2}=a b \ldots \ldots \text { (2) } \end{array}$

Considering second curve

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

Substituting this in eq (1)

\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{a b-x^{2}}{b^{2}}=1 \\ &\frac{x^{2} b^{2}+a^{2\left(a b-x^{2}\right)}}{a^{2} b^{2}}=1 \\ &x^{2} b^{2}+a^{3} b-a^{2} x^{2}=a^{2} b^{2} \end{aligned}

\begin{aligned} &x^{2} b^{2}-a^{2} x^{2}=a^{2} b^{2}-a^{3} b \\ &x^{2}\left(b^{2}-a^{2}\right)=a^{2} b(b-a) \\ &x^{2}=\frac{a^{2} b(b-a)}{b^{2}-a^{2}}::\left(x^{2}-y^{2}\right)=(x+y)(x-y) \end{aligned}

\begin{aligned} x^{2} &=\frac{a^{2} b(b-a)}{(b+a)(b-a)} \\ x^{2} &=\frac{a^{2} b}{(b+a)} \\ x &=\pm \sqrt{\frac{a^{2} b}{(b+a)}} \ldots \ldots \text { (3) } \end{aligned}

Since,   \begin{aligned} y^{2}=ab-x^{2} \end{aligned}

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

\begin{aligned} y^{2} &=\frac{a b^{2}}{b+a} \\ y &=\pm \sqrt{\frac{a b^{2}}{b+a}} \ldots \ldots \text { (4) } \end{aligned}

Since curves are        $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \& x^{2}+y^{2}=a b$

Differentiating above with respect to x,

\begin{aligned} &=\frac{2 x^{2}}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0 \\ &=\frac{y}{b^{2}} \cdot \frac{d y}{d x}=\frac{-x}{a^{2}} \end{aligned}

\begin{aligned} &=\frac{d y}{d x}=\frac{-x}{a^{2}} \times \frac{b^{2}}{y}=\frac{-b^{2} x}{a^{2} y}\\ &=m_{1}=\frac{d y}{d x}=\frac{-b^{2} x}{a^{2} y} \ldots \ldots(5) \end{aligned}

Second curve is  \begin{aligned} & x^{2} +y^{2}=ab \end{aligned}
\begin{aligned} &=2 x^{2}+2 y^{2} \cdot \frac{d y}{d x}=0 \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{y} \ldots \ldots \text { (6) } \end{aligned}

Substituting (3) in (4) above values for   m1 &  m2, we get

At$\left(\sqrt{\frac{a^{2} b}{b+a}}, \sqrt{\frac{a b^{2}}{b+a}}\right)$ in eq (6), we get
\begin{aligned} &=\frac{d y}{d x}=\frac{-\sqrt{\frac{a^{2} b}{b+a}}}{\sqrt{\frac{a b^{2}}{a+b}}} \\ &=\frac{d y}{d x}=\frac{-a \sqrt{\frac{b}{b+a}}}{b \sqrt{\frac{a}{b+a}}} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-a \sqrt{b}}{b \sqrt{a}} \\ &=m_{2}=\frac{d y}{d x}=-\sqrt{\frac{a}{b}} \end{aligned}

When  $m_{1}=\frac{-b \sqrt{b}}{a \sqrt{a}}$ and  $m_{2}=-\sqrt{\frac{a}{b}}$

Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{-b \sqrt{b}}{a \sqrt{a}}--\sqrt{\frac{a}{b}}\left(-\sqrt{\frac{a}{b}}\right)}{1+\left(\frac{-b \sqrt{b}}{a \sqrt{a}}\right)\left(-\sqrt{\frac{a}{b}}\right)}\right| \end{aligned}
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{\frac{-b \sqrt{b} \times \sqrt{b}+a \sqrt{a} \times \sqrt{a}}{a \sqrt{a} \times \sqrt{b}}}{1+\frac{b}{a}}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{-b \times b+a \times a}{a \sqrt{a b}}}{1+\frac{b}{a}}\right| \end{aligned}

\begin{aligned} &\operatorname{tan} \theta=\left|\frac{\frac{a^{2}-b^{2}}{a \sqrt{a b}}}{\frac{a+b}{a}}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{(a-b)(a+b)}{\sqrt{a b}}}{a+b}\right| \end{aligned}

\begin{aligned} &\operatorname{tan} \theta=\left|\frac{a-b}{\sqrt{a b}}\right| \\ &\operatorname{tan} \theta=\frac{a-b}{\sqrt{a b}} \\ &\theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right) \end{aligned}