# Equation of the Tangent to the Conic

The equation of the tangent to the conic $a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$ at the point $\left( {{x_1},{y_1}} \right)$ can be written in the form

$ax{x_1} + by{y_1} + h\left( {x{y_1} + {x_1}y} \right) + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$

Proof:
Since the point $\left( {{x_1},{y_1}} \right)$ lies on the conic
$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

So the above equation (i) becomes
$a{x_1}^2 + b{y_1}^2 + 2h{x_1}{y_1} + 2g{x_1} + 2f{y_1} + c = 0\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Now differentiating equation (i) of a general conic with respect to $x$, we have
$\begin{gathered} \frac{{dy}}{{dx}} = – \frac{{ax + hy + g}}{{hx + by + f}} \\ \Rightarrow {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},{y_1}} \right)}} = – \frac{{a{x_1} + h{y_1} + g}}{{h{x_1} + b{y_1} + f}} \\ \end{gathered}$

The equation of the tangent at the point $\left( {{x_1},{y_1}} \right)$ is
$\begin{gathered} y – {y_1} = – \frac{{a{x_1} + h{y_1} + g}}{{h{x_1} + b{y_1} + f}}\left( {x – {x_1}} \right) \\ \Rightarrow ax{x_1} + by{y_1} + h\left( {x{y_1} + {x_1}y} \right) + gx + fy = a{x_1}^2 + b{y_1}^2 + 2h{x_1}{y_1} + g{x_1} + f{y_1} \\ \end{gathered}$

Adding $g{x_1} + f{y_1} + c$ on both sides and using equation (ii), we have
$ax{x_1} + by{y_1} + h\left( {x{y_1} + {x_1}y} \right) + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$

NOTE: The above theorem suggests that the equation of the tangent to the conic at the point $\left( {{x_1},{y_1}} \right)$ may be obtained by replacing ${x^2}$ by $x{x_1}$; ${y^2}$ by $y{y_1}$; $2xy$ by $x{y_1} + {x_1}y$; $2x$ by $x + {x_1}$; $2y$ by $y + {y_1}$. Keeping these replacements in mind, we have the following conclusions:

(i) For parabola ${y^2} = 4ax$, the equation of the tangent is $y{y_1} = 2a\left( {x + {x_1}} \right)$

(ii) For ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, the equation of the tangent is $\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1$

(iii) For hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$, the equation of the tangent is $\frac{{x{x_1}}}{{{a^2}}} – \frac{{y{y_1}}}{{{b^2}}} = 1$