Question

Find the condition for the following set of curves to intersect orthogonally:

(i) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\mathrm{and}xy=c^2$
(ii) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\mathrm{and}\frac{x^2}{{\mathrm{A}}^2}-\frac{y^2}{{\mathrm{B}}^2}=1$

Answer 1

$$\begin{gathered} \left(i\right)\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(1\right) \\ xy=c^2...\left(2\right) \\ \text{Let the curves intersect orthogonally at}\left(x_1,y_1\right)\text{.} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (1) on both sides w.r.t.}x,\mathrm{we}\mathrm{get} \\ \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{x{b}^2}{a^{2}y} \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{x_1{b}^2}{a^2{y}_1} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (2) on both sides w.r.t.}x,\mathrm{we}\mathrm{get} \\ x\frac{dy}{dx}+y=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-y}{x} \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{-y_1}{x_1} \\ \text{It is given that the curves intersect orhtogonally at}\left(x_1,y_1\right)\text{.} \\ \therefore m_1\times m_2=-1 \\ \Rightarrow \frac{x_1{b}^2}{a^2{y}_1}\times \frac{-y_1}{x_1}=-1 \\ \Rightarrow a^2=b^2 \end{gathered}$$

(ii) The condition for the curves $a{x}^2+b{y}^2=1\text{and}a\text{'}{x}^2+b\text{'}{y}^2=1$ to intersect orthogonally is given below:
$$\begin{gathered} \frac{1}{a}-\frac{1}{b}=\frac{1}{a\text{'}}-\frac{1}{b\text{'}} \\ \text{So, the condition for the curves}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{and}\frac{x^2}{A^2}-\frac{y^2}{B^2}=1\text{to intersect orthogonally is} \\ \frac{1}{{\displaystyle \frac{1}{a^2}}}-\frac{1}{{\displaystyle \frac{1}{b^2}}}=\frac{1}{{\displaystyle \frac{1}{A^2}}}-\frac{1}{{\displaystyle \frac{-1}{B^{\mathbf{2}}}}} \\ \Rightarrow a^2-b^2=A^2+B^2 \end{gathered}$$