Question

Show that the following curves intersect orthogonally at the indicated points:

(i) x² = 4y and 4y + x² = 8 at (2, 1)
(ii) x² = y and x³ + 6y = 7 at (1, 1)
(iii) y² = 8x and 2x² + y² = 10 at $\left(1,2\sqrt{2}\right)$

Answer 1

$$\begin{gathered} \left(i\right)x^2=4y...\left(1\right) \\ 4y+x^2=8...\left(2\right) \\ \text{Given point is}\left(2,1\right) \\ \text{Differentiating (1) w.r.t.}\mathit{\text{x,}} \\ 2x=4\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{x}{2} \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(2,1\right)}=\frac{2}{2}=1 \\ \text{Differenntiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}} \\ 4\frac{dy}{dx}+\mathit{2}x=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x}{2} \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(2,1\right)}=\frac{-2}{2}=-1 \\ \mathrm{Since},m_1\times m_2=-1 \end{gathered}$$
Hence, the given curves intersect orthogonally at the given point.

$$\begin{gathered} \left(ii\right)x^2=y...\left(1\right) \\ {x}^3+6y=7...\left(2\right) \\ \text{Given point is}\left(1,1\right) \\ \text{Differentiating (1) w.r.t.}\mathit{\text{x,}} \\ 2x=\frac{dy}{dx} \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=2\left(1\right)=2 \\ \text{Differenntiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}} \\ 3x^2+6\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x^2}{2} \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=\frac{-1}{2} \\ \mathrm{Since},m_1\times m_2=-1 \end{gathered}$$
Hence, the given curves intersect orthogonally at the given point.

$$\begin{gathered} \left(iii\right)y^2=8x...\left(1\right) \\ 2x^2+y^2=10...\left(2\right) \\ \text{Given point is}\left(1,2\sqrt{2}\right) \\ \text{Differentiating (1) w.r.t.}\mathit{\text{x,}} \\ 2y\frac{dy}{dx}=8 \\ \Rightarrow \frac{dy}{dx}=\frac{4}{y} \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(1,2\sqrt{2}\right)}=\frac{4}{2\sqrt{2}}=\sqrt{2} \\ \text{Differenntiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}} \\ 4x+2y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{y} \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(1,2\sqrt{2}\right)}=\frac{-2}{2\sqrt{2}}=\frac{-1}{\sqrt{2}} \\ \mathrm{Since},m_1\times m_2=-1 \end{gathered}$$
Hence, the given curves intersect orthogonally at the given point.