Question

Show that the following set of curves intersect orthogonally.

(i) y = x³ and 6y = 7 − x²
(ii) x³ − 3xy² = −2 and 3x²y − y³ = 2
(iii) x² + 4y² = 8 and x² − 2y² = 4

Answer 1

$$\begin{gathered} \left(i\right)y=x^3...\left(1\right) \\ 6y=7-x^2...\left(2\right) \\ \text{From (1) and (2) we get} \\ 6x^3=7-x^2 \\ \Rightarrow 6x^3+x^2-7=0 \\ \mathit{\text{x}}\text{=1 satisfies this.} \\ \text{Dividing this by}\mathit{\text{x}}\text{-1 ,we get} \\ 6x^2+7x+7=0, \\ {\text{Discriminant = 7}}^2\text{-4}\left(6\right)\left(7\right)\text{=-119<0} \\ \text{So, this has no real roots.} \\ \text{When}\mathit{\text{x}}\text{=1,}\mathit{\text{y}}\text{=}{\mathit{\text{x}}}^3\text{=1 (From (1))} \\ \mathrm{So},\left(x,y\right)\text{=}\left(1,1\right) \\ \text{Differentiating (1) w.r.t.}\mathit{\text{x,}} \\ \frac{dy}{dx}=3x^2 \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=3 \\ \text{Differenntiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}} \\ \mathit{6}\frac{dx}{dx}=\mathit{-}\mathit{2}x \\ \Rightarrow \frac{dx}{dx}=\frac{\mathit{-}\mathit{2}x}{6}=\frac{-x}{3} \\ \Rightarrow m_{\mathit{2}}={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=\frac{-1}{3} \\ \mathrm{Now},m_1\times m_2=3\times \frac{-1}{3} \\ \Rightarrow m_1\times m_2=-1 \\ \mathrm{Since},m_1\times m_2=-1 \end{gathered}$$
So, the given curves intersect orthogonally.

$$\begin{gathered} \left(ii\right)\text{Let the given curves intersect at}\left(x_1,y_1\right) \\ {x}^3-3x{y}^2=-2...\left(1\right) \\ 3x^{2}y-y^3=2...\left(2\right) \\ \text{Differentiating (1) w.r.t.}\mathit{\text{x,}} \\ 3x^{\mathit{2}}\mathit{-}3y^{\mathit{2}}\mathit{-}6xy\frac{dy}{dx}=0 \\ \mathit{\Rightarrow }\frac{dy}{dx}=\frac{3x^{\mathit{2}}\mathit{-}3y^{\mathit{2}}}{6xy}=\frac{x^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}}{2xy} \\ \mathit{\Rightarrow }{m}_{\mathit{1}}={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{{x_1}^{\mathit{2}}\mathit{-}{y_1}^{\mathit{2}}}{2x_1{y}_1} \\ \text{Differenntiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}} \\ 3x^{\mathit{2}}\frac{dy}{dx}\mathit{+}6xy\mathit{-}3y^{\mathit{2}}\frac{dy}{dx}\mathit{=}\mathit{0} \\ \mathit{\Rightarrow }\frac{dy}{dx}\left(3x^{\mathit{2}}\mathit{-}3y^{\mathit{2}}\right)=\mathit{-}6xy \\ \mathit{\Rightarrow }\frac{dy}{dx}=\frac{\mathit{-}6xy}{3x^{\mathit{2}}\mathit{-}3y^{\mathit{2}}}=\frac{\mathit{-}2xy}{x^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}} \\ \mathit{\Rightarrow }{m}_2={\left(\frac{dy}{dx}\right)}_{\left(x_{\mathit{1}}\mathit{,}\mathit{}{y}_{\mathit{1}}\right)}\mathit{=}\frac{\mathit{-}2x_1{y}_1}{{x_1}^{\mathit{2}}\mathit{-}{y_1}^{\mathit{2}}} \\ \mathrm{Now},m_1\times m_2=\frac{{x_1}^{\mathit{2}}\mathit{-}{y_1}^{\mathit{2}}}{2x_1{y}_1}\times \frac{\mathit{-}2x_1{y}_1}{{x_1}^{\mathit{2}}\mathit{-}{y_1}^{\mathit{2}}} \\ \Rightarrow m_1\times m_2=-1 \\ \mathrm{Since},m_1\times m_2=-1 \end{gathered}$$
So, the given curves intersect orthogonally.

$$\begin{gathered} \left(\mathrm{iii}\right)x^2+4y^2=8...\left(1\right) \\ {x}^2-2y^2=4...\left(2\right) \\ \text{From (1) and (2) we get} \\ 6y^2=4 \\ \Rightarrow y^2=\frac{2}{3} \\ \Rightarrow y=\frac{\sqrt{2}}{\sqrt{3}}\text{or}y=\frac{-\sqrt{2}}{\sqrt{3}} \\ \text{From (1),} \\ {x}^2+\frac{8}{3}=8 \\ \Rightarrow x^2=\frac{16}{3} \\ \Rightarrow x=\pm \frac{4}{\sqrt{3}} \\ \text{So,}\left(x,y\right)\text{=}\left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right)\text{,}\left(\frac{4}{\sqrt{3}},\frac{-\sqrt{2}}{\sqrt{3}}\right)\text{,}\left(\frac{-4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right)\text{,}\left(\frac{-4}{\sqrt{3}},-\frac{\sqrt{2}}{\sqrt{3}}\right) \\ \text{Consider point}\left(x_1,y_1\right)\text{=}\left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right) \\ \text{Differentiating (1) w.r.t.}\mathit{\text{x,}} \\ 2x+8y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}\mathit{=}\frac{\mathit{-}x}{\mathit{4}y} \\ \Rightarrow m_{\mathit{1}}\mathit{=}{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}_{\left(\frac{\mathit{4}}{\sqrt{\mathit{3}}}\mathit{,}\mathit{}\frac{\sqrt{\mathit{2}}}{\sqrt{\mathit{3}}}\right)}\mathit{=}\frac{-\frac{4}{\sqrt{3}}}{4\frac{\sqrt{2}}{\sqrt{3}}}=\frac{-1}{\sqrt{2}} \\ \text{Differenntiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}} \\ \mathit{2}x\mathit{-}\mathit{4}y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{x}{\mathit{2}y} \\ \Rightarrow m_{\mathit{2}}={\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}_{\left(\frac{\mathit{4}}{\sqrt{\mathit{3}}}\mathit{,}\mathit{}\frac{\sqrt{\mathit{2}}}{\sqrt{\mathit{3}}}\right)}\mathit{=}\frac{\frac{4}{\sqrt{3}}}{2\frac{\sqrt{2}}{\sqrt{3}}}=\sqrt{2} \\ \mathrm{Now},m_1\times m_2=\frac{-1}{\sqrt{2}}\times \sqrt{2} \\ \Rightarrow \mathit{}{m}_1\times m_2=-1 \\ \mathrm{Since},m_1\times m_2=-1 \\ \text{Hence,, the curves are orthogonal at}\left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right)\text{.} \\ \text{Similarly, we can see that the curves are orthogonal in each possibility of}\left(x_1,y_1\right)\text{.} \end{gathered}$$