Equation of Tangent and Normal to the Ellipse
The equations of tangent and normal to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ at the point $$\left( {{x_1},{y_1}} \right)$$ are $$\frac{{{x_1}x}}{{{a^2}}} + \frac{{{y_1}y}}{{{b^2}}} = 1$$ and $${a^2}{y_1}x – {b^2}{x_1}y – \left( {{a^2} – {b^2}} \right){x_1}{y_1} = 0$$ respectively.
Consider that the standard equation of ellipse with vertex at origin $$\left( {0,0} \right)$$ can be written as
\[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]
Since the point $$\left( {{x_1},{y_1}} \right)$$ lies on the given ellipse, it must satisfy equation (i). So we have
\[\frac{{{x_1}^2}}{{{a^2}}} + \frac{{{y_1}^2}}{{{b^2}}} = 1\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)\]
Now differentiating equation (i) on both sides with respect to $$x$$, we have
\[\begin{gathered} \frac{{2x}}{{{a^2}}} + \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0 \Rightarrow \frac{y}{{{b^2}}}\frac{{dy}}{{dx}} = – \frac{x}{{{a^2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{{{b^2}x}}{{{a^2}y}} \\ \end{gathered} \]
If $$m$$ represents the slope of the tangent at the given point $$\left( {{x_1},{y_1}} \right)$$, then
\[m = {\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = – \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}\]
The equation of a tangent at the given point $$\left( {{x_1},{y_1}} \right)$$ is
\[\begin{gathered} y – {y_1} = – \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}\left( {x – {x_1}} \right) \\ \frac{{{y_1}}}{{{b^2}}}\left( {y – {y_1}} \right) = – \frac{{{x_1}}}{{{a^2}}}\left( {x – {x_1}} \right) \\ \Rightarrow \frac{{{y_1}y}}{{{b^2}}} – \frac{{{y_1}^2}}{{{b^2}}} = – \frac{{{x_1}x}}{{{a^2}}} + \frac{{{x_1}^2}}{{{a^2}}} \\ \Rightarrow \frac{{{x_1}x}}{{{a^2}}} + \frac{{{y_1}y}}{{{b^2}}} = \frac{{{x_1}^2}}{{{a^2}}} + \frac{{{y_1}^2}}{{{b^2}}} \\ \Rightarrow \boxed{\frac{{{x_1}x}}{{{a^2}}} + \frac{{{y_1}y}}{{{b^2}}} = 1} \\ \end{gathered} \]
This is the equation of the tangent to the given ellipse at $$\left( {{x_1},{y_1}} \right)$$.
The slope of the normal at $$\left( {{x_1},{y_1}} \right)$$ is $$ – \frac{1}{m} = – \left( { – \frac{{{a^2}{x_1}}}{{{b^2}{y_1}}}} \right) = \frac{{{a^2}{x_1}}}{{{b^2}{y_1}}}$$
The equation of the normal at the point $$\left( {{x_1},{y_1}} \right)$$ is $$y – {y_1} = \frac{{{a^2}{x_1}}}{{{b^2}{y_1}}}\left( {x – {x_1}} \right)$$
\[\begin{gathered} \Rightarrow {a^2}{y_1}x – {b^2}{x_1}y – {a^2}{x_1}{y_1} + {b^2}{x_1}{y_1} = 0 \\ \Rightarrow \boxed{{a^2}{y_1}x – {b^2}{x_1}y – \left( {{a^2} – {b^2}} \right){x_1}{y_1} = 0} \\ \end{gathered} \]
This is the equation of the normal to the given ellipse at $$\left( {{x_1},{y_1}} \right)$$.
Artem Piatkov
February 1 @ 2:07 am
Hello. There is probably an error in the expression for the normal at (x1,y1).
It should be (a^2 * y1) / (b^2 * x1).
The following expressions should be also corrected accordingly.
fery
March 17 @ 12:25 am
If we draw 2 tangents to an ellipse from a single point. Will the bisector of the angel between the two tangents go through the origin?