Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 154: 52

For $x=1\pm\sqrt2$, $\dfrac{dx}{dy}$ is equal to zero. Since $\dfrac{dx}{dy}$ is equal to zero only when the tangent line is vertical. The tangent lines at $x = 1 ±\sqrt2$ to the conchoid with equation $(x − 1)^2(x^2 + y^2) = 2x^2$ are vertical.

Firstly differentiate $(x − 1)^2(x^2 + y^2) = 2x^2$ with respect to $y$ using product rule and chain rule. $\dfrac{d}{dy}\left((x-1)^2(x^2+y^2)\right)=\dfrac{d}{dy}\left(2x^2\right)$ $\implies (x-1)^2\dfrac{d}{dy}\left((x^2+y^2)\right)+(x^2+y^2)\dfrac{d}{dy}\left((x-1)^2\right)=4x\dfrac{dx}{dy}$ $\implies (x-1)^2\left(2x\dfrac{dx}{dy}+2y\right)+(x^2+y^2)\times2(x-1)\dfrac{d}{dy}\left(x-1\right)=4x\dfrac{dx}{dy}$ $\implies (x-1)^2\left(2x\dfrac{dx}{dy}+2y\right)+2(x^2+y^2)(x-1)\dfrac{dx}{dy}=4x\dfrac{dx}{dy}$ $\implies 2x(x-1)^2\dfrac{dx}{dy}+2y(x-1)^2+2(x^2+y^2)(x-1)\dfrac{dx}{dy}=4x\dfrac{dx}{dy}$ Now isolate terms with $\dfrac{dx}{dy}$. We get, $2x(x-1)^2\dfrac{dx}{dy}-4x\dfrac{dx}{dy}+2(x^2+y^2)(x-1)\dfrac{dx}{dy}=-2y(x-1)^2$ Now take $\dfrac{dx}{dy}$ common. $\left(2x(x-1)^2-4x+2(x^2+y^2)(x-1)\right)\dfrac{dx}{dy}=-2y(x-1)^2$ $\implies \dfrac{dx}{dy}=\dfrac{-2y(x-1)^2}{2x(x-1)^2-4x+2(x^2+y^2)(x-1)}$ Now substitute $x=1+\sqrt2$ in $(x − 1)^2(x^2 + y^2) = 2x^2$ and then solve for $y$. $(1+\sqrt2 − 1)^2((1+\sqrt2)^2 + y^2) = 2(1+\sqrt2)^2$ $\implies (\sqrt2 )^2(1+2+2\sqrt2 + y^2) = 2(1+2+2\sqrt2)$ $\implies 2(3+2\sqrt2 + y^2) = 2(3+2\sqrt2)$ $\implies 3+2\sqrt2 + y^2 = 3+2\sqrt2$ $\implies y^2 = 0$ or $y=0$ Now substitute $x=1-\sqrt2$ in $(x − 1)^2(x^2 + y^2) = 2x^2$ and then solve for $y$. $(1-\sqrt2 − 1)^2((1-\sqrt2)^2 + y^2) = 2(1-\sqrt2)^2$ $\implies (-\sqrt2 )^2(1+2-2\sqrt2 + y^2) = 2(1+2-2\sqrt2)$ $\implies 2(3-2\sqrt2 + y^2) = 2(3-2\sqrt2)$ $\implies 3-2\sqrt2 + y^2 = 3-2\sqrt2$ $\implies y^2 = 0$ or $y=0$ If we substitute $y=0$ in $\dfrac{dx}{dy}=\dfrac{-2y(x-1)^2}{2x(x-1)^2-4x+2(x^2+y^2)(x-1)}$. We get, $\dfrac{dx}{dy}=0$. Since $y=0$ when $x=1\pm\sqrt2$ Thus, for $x=1\pm\sqrt2$, $\dfrac{dx}{dy}=0$. Since $\dfrac{dx}{dy}$ is equal to zero only when the tangent line is vertical. The tangent lines at $x = 1 ±\sqrt2$ to the conchoid with equation $(x − 1)^2(x^2 + y^2) = 2x^2$ are vertical.