Tangent Line Formula for a Conic Section

The tangent line formula (or splitting rule) is a method in analytic geometry that allows us to determine the equation of the tangent line to a conic section in a Cartesian plane at the point \( P(x_0, y_0) \). The formula is: $$ a_{11}x_0 x + a_{22}y_0 y + a_{12}(x y_0 + x_0 y) + \\ + a_{01}(x + x_0) + a_{02}(y + y_0) + a_{00} = 0 $$ The coefficients are taken from the conic section's equation:
$ a_{11}x^2 + a_{22}y^2 + 2a_{12}xy + 2a_{01}x + 2a_{02}y + a_{00} = 0 $

In other words, the tangent line formula allows us to find the equation of the tangent line by appropriately transforming the terms of the original conic equation without using derivatives.

A conic section is described by the general equation:

$$ a_{11}x^2 + a_{22}y^2 + 2a_{12}xy + 2a_{01}x + 2a_{02}y + a_{00} = 0 $$

The splitting rule states that the tangent line equation at \( P(x_0, y_0) \) is obtained by making the following formal substitutions in the variables of the conic's equation:

Substitutions for the second-degree terms:

  • \( x^2 \rightarrow x_0 x \)
  • \( y^2 \rightarrow y_0 y \)
  • \( xy \rightarrow \frac{1}{2}(x y_0 + x_0 y) \)

Substitutions for the first-degree terms:

  • \( x \rightarrow \frac{1}{2}(x + x_0) \)
  • \( y \rightarrow \frac{1}{2}(y + y_0) \)

Applying these substitutions to the conic's equation gives the equation of the tangent line at the point \( P(x_0, y_0) \):

$$ a_{11}x_0 x + a_{22}y_0 y + a_{12}(x y_0 + x_0 y) + a_{01}(x + x_0) + a_{02}(y + y_0) + a_{00} = 0 $$

Note. Remembering the tangent line formula for a conic section can be quite difficult. One way to memorize it is to perform the formal substitutions directly on the conic's equation. Start from the conic's equation you already know and substitute the quadratic terms with $$ x^2 \rightarrow x_0 x $$ $$ y^2 \rightarrow y_0 y $$ Then substitute \( xy \) with their average $$ xy \rightarrow \frac{1}{2}(x y_0 + x_0 y) $$ Finally, replace the linear terms \( x \) and \( y \) with the following averages $$ x \rightarrow \frac{1}{2}(x + x_0) $$ $$ y \rightarrow \frac{1}{2}(y + y_0) $$ where \( x_0 \) and \( y_0 \) are the coordinates of the point of tangency. This mnemonic trick helps to remember the splitting formula more easily.

    A Practical Example

    Here's a practical example using the splitting formula to find the equation of the tangent line to a conic section.

    Consider an ellipse given by the equation:

    $$ -2x^2 - 3y^2 + 2x = -12 $$

    Let's rewrite it in general form:

    $$ -2x^2 - 3y^2 + 2x + 12 = 0 $$

    We want to find the equation of the tangent line to this ellipse at the point \( P(1, 2) \).

    $$ -2x^2 - 3y^2 + 2x + 12 = 0 $$

    We apply the formal substitutions to the terms of the conic's equation.

    Substitutions for the second-degree terms:

    $$ x^2 \rightarrow x_0 x $$

    $$ y^2 \rightarrow y_0 y  $$

    Thus, the conic's equation becomes:

    $$ -2x_0 x - 3y_0 y + 2x + 12 = 0 $$

    Next, substitute the linear term \( x \rightarrow \frac{1}{2}(x + x_0) \)

    $$ -2x_0 x - 3y_0 y + 2 \cdot \frac{1}{2}(x + x_0) + 12 = 0 $$

    $$ -2x_0 x - 3y_0 y + (x + x_0) + 12 = 0 $$

    $$ -2x_0 x - 3y_0 y + x + x_0 + 12 = 0 $$

    The point of tangency we're looking for is at coordinates \( P(1, 2) \), which means \( x_0 = 1 \) and \( y_0 = 2 \).

    $$ -2 \cdot 1 \cdot x - 3 \cdot 2 \cdot y + x + 1 + 12 = 0 $$

    $$ -x - 6y + 13 = 0 $$

    Simplify further by multiplying both sides of the equation by -1:

    $$ x + 6y - 13 = 0 $$

    The final result is the equation of the tangent line at the point \( P(1,2) \).

    an example of the tangent line at the point (1,2) of the conic section

    Alternative Solution

    To find the tangent line to the conic section at the point \( P(1,2) \), you can also follow this alternative procedure.

    $$ -2x^2 - 3y^2 + 2x + 12 = 0 $$

    Knowing that in general the equation of a conic section is:

    $ a_{11}x^2 + a_{22}y^2 + 2a_{12}xy + 2a_{01}x + 2a_{02}y + a_{00} = 0 $

    We identify the coefficients of the conic $ -2x^2 - 3y^2 + 2x + 12 = 0 $, which in this case are:

    •  \( a_{11} = -2 \)
    •  \( a_{22} = -3 \)
    •  \( a_{12} = 0 \) (since there is no \( xy \) term)
    •  \( a_{01} = 1 \) (since $ 2a_{01}x = 2 $, so $ a_{01}=\frac{1}{2} \cdot 2 = 1 $)
    •  \( a_{02} = 0 \) (since there is no \( y \) term)
    •  \( a_{00} = -12 \)

    Now we write the tangent line formula for the conic section.

    $$ a_{11}x_0 x + a_{22}y_0 y + a_{12}(x y_0 + x_0 y) + a_{01}(x + x_0) + a_{02}(y + y_0) + a_{00} = 0 $$

    We substitute the coefficients we just found.

    $$ -2x_0 x - 3y_0 y + 0 \cdot (x y_0 + x_0 y) + 1 \cdot (x + x_0) + 0 \cdot (y + y_0) + 12 = 0 $$

    $$ -2x_0 x - 3y_0 y + x + x_0 + 12 = 0 $$

    In this case, the point of tangency is at coordinates \( P(1, 2) \), so \( x_0 = 1 \) and \( y_0 = 2 \).

    $$ -2 \cdot 1 \cdot x - 3 \cdot 2 \cdot y + x + 1 + 12 = 0 $$

    $$ -x - 6y + 13 = 0 $$

    $$ x + 6y - 13 = 0 $$

    This is the equation of the tangent line we were looking for.

    So, we have found the tangent line equation to the conic section at the point \( P(1,2) \).

    an example of the tangent line at the point (1,2) of the conic section

    And that's how it's done.

     
     

    Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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