Find the tangents at an arbitrary point (x_0,y_0) to 1°

Find the tangents at an arbitrary point $\bigl(x_{0},y_{0}\bigr)$ to 1° $x^{2}+y^{2} = a^{2}$; 2° $y^{2} = 2px$; 3° $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} = 1$; 4° $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}} = 1$.

Step 1: The method

For $F(x,y) = 0$, implicit differentiation gives $y' = -\dfrac{F_{x}}{F_{y}}$, and the tangent at $\bigl(x_{0},y_{0}\bigr)$ is $y-y_{0} = y'\bigl(x_{0},y_{0}\bigr)\bigl(x-x_{0}\bigr)$. In each case below the equation of the curve is then used to simplify the right-hand side.

Step 2: 1° — the circle

From $2x+2yy' = 0$, $y' = -\dfrac{x}{y}$, so

\[ y_{0}\bigl(y-y_{0}\bigr) = -x_{0}\bigl(x-x_{0}\bigr) \;\Longrightarrow\; x_{0}x+y_{0}y = x_{0}^{2}+y_{0}^{2} = a^{2}. \]

Step 3: 2° — the parabola

From $2yy' = 2p$, $y' = \dfrac{p}{y}$, so $y_{0}y-y_{0}^{2} = px-px_{0}$; using $y_{0}^{2} = 2px_{0}$, $y_{0}y = px+px_{0} = p\bigl(x+x_{0}\bigr)$.

Step 4: 3° — the ellipse

From $\dfrac{2x}{a^{2}}+\dfrac{2y}{b^{2}}y' = 0$, $y' = -\dfrac{b^{2}x}{a^{2}y}$, so $b^{2}x_{0}x+a^{2}y_{0}y = b^{2}x_{0}^{2}+a^{2}y_{0}^{2}$; dividing by $a^{2}b^{2}$, \[\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^{2}} = \frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}} = 1.\]

Step 5: 4° — the hyperbola

The same computation with a minus sign gives $y' = \dfrac{b^{2}x}{a^{2}y}$ and

\[ \frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}} = \frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}} = 1. \]

Step 6: Verification

\[\text{Each formula reduces to the curve's own equation when } \bigl(x,y\bigr) = \bigl(x_{0},y_{0}\bigr) \;\;\checkmark\]

\[\text{Circle, } \bigl(x_{0},y_{0}\bigr) = \bigl(a,0\bigr):\ x = a,\ \text{the vertical tangent} \;\;\checkmark\]

\[\text{Parabola, } \bigl(x_{0},y_{0}\bigr) = \bigl(0,0\bigr):\ 0 = px,\ \text{i.e. the } y\text{-axis} \;\;\checkmark\]