Tangent Line to Conic Sections

Concept of Tangent Lines

Have you ever seen a basketball that touches the rim perfectly? At that point of contact, the ball just touches one point without going through the rim. This concept is what we call a tangent line in mathematics!

A tangent line to a conic section is a line that touches the curve at exactly one point only. Unlike a secant line that intersects the curve at two points, a tangent line only touches at one point and doesn't cut through the curve at all.

Just imagine if you have a parabola $y = x^2$. A tangent line will touch the parabola at one specific point, while a secant line will intersect the parabola at two different points.

For conic sections like parabolas, ellipses, and hyperbolas, there are several ways to determine the equation of their tangent lines depending on the information we have.

Point on the Curve

If we already know where the point of tangency is, determining the tangent line becomes really easy! The basic concept uses the division principle which is practical for conic sections.

Division Principle

To determine the tangent line through point $T(x_1, y_1)$ on a conic section, we can use the division principle. The method is simple, namely divide each squared term into a linear term at the point of tangency.

For example, if we have a parabola $y^2 = 4px$ and tangent point $T(x_1, y_1)$, then the tangent line equation becomes:

Let's take an example of parabola $y^2 = 8x$ at point $(2, 4)$.

From parabola $y^2 = 8x$, we know $4p = 8$ so $p = 2$. The tangent line at point $(2, 4)$ is:

This division principle makes calculations much easier! We don't need to bother calculating derivatives. Just "divide" each squared term into multiplication with the tangent point coordinates!

Specific Gradient

Sometimes we don't know the tangent point, but we know the slope or gradient of the tangent line. In cases like this, we substitute the line equation with gradient $m$ into the conic section equation.

For example, we want to find the tangent line of hyperbola $4x^2 - 9y^2 = 36$ that is perpendicular to line $x + 4y + 10 = 0$.

First step, we determine the gradient of the tangent line. Since the tangent line is perpendicular to $x + 4y + 10 = 0$, then the original line's gradient $m_g = -\frac{1}{4}$ so the tangent line's gradient $m_s = 4$.

The tangent line equation of a hyperbola with gradient $m$ is $y = mx + c$. We substitute it into the hyperbola equation:

For a tangent line, the discriminant must be zero:

So we get $c = 0$ and the tangent line equation is $y = 4x$.

Point Outside the Curve

If the point is outside the conic section, we can have two tangent lines that can be drawn from that point to the curve. The concept is similar to drawing lines from a point outside a circle.

For example, take parabola $y^2 = 8x$ with point $A(2, 5)$. From point $A$, we can draw two different tangent lines to the parabola.

To determine the tangent line equation through an external point, we use the polar or pole equation. For parabola $y^2 = 8x$ with point $A(2,5)$, the polar equation is:

Substitute $x_1 = 2$ and $y_1 = 5$:

Now we substitute into the parabola equation to find the tangent points:

Using the quadratic formula, we get $x_1 = \frac{1}{2}$ and $x_2 = 8$. So the tangent points are at $(\frac{1}{2}, 2)$ and $(8, 8)$.

Formula Summary

Here's the complete table of conic section equations and their tangent line equations. Keep this well, because you'll use it frequently!

All these formulas are based on the division principle that makes calculations easier! No need to mess around with calculus.

Exercises

1. Find the tangent line equation of parabola $y^2 = 12x$ at point $(3, 6)$.

2. There's a hyperbola $4x^2 - 9y^2 = 36$. Find the tangent line equation that is perpendicular to line $x + 4y + 10 = 0$.

3. Find the tangent line equation of ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ that passes through point $A(1, 3)$.

4. There's a parabola $y^2 = 8x$. Find the tangent line equation of this parabola that passes through point $A(2, 5)$.

Answer Key

1. Answer:

   Given: parabola $y^2 = 12x$ with tangent point $(3, 6)$.

   From equation $y^2 = 12x$, we get $4p = 12$ so $p = 3$.

   Use the parabola tangent line formula:

   $$yy_1 = 2p(x + x_1)$$

   $$y \cdot 6 = 2 \cdot 3 \cdot (x + 3)$$

   $$6y = 6(x + 3)$$

   $$6y = 6x + 18$$

   $$y = x + 3$$

   So the tangent line equation is $y = x + 3$.

2. Answer:

   Given: hyperbola $4x^2 - 9y^2 = 36$ or $\frac{x^2}{9} - \frac{y^2}{4} = 1$.

   Line $x + 4y + 10 = 0$ has gradient $m = -\frac{1}{4}$.

   Since the tangent line is perpendicular to that line, then the tangent line gradient is:

   $$m_s = -\frac{1}{m} = -\frac{1}{-\frac{1}{4}} = 4$$

   Hyperbola tangent line formula with gradient $m$:

   $$y = mx \pm \sqrt{a^2m^2 - b^2}$$

   With $a^2 = 9$, $b^2 = 4$, and $m = 4$:

   $$y = 4x \pm \sqrt{9 \cdot 16 - 4}$$

   $$y = 4x \pm \sqrt{144 - 4}$$

   $$y = 4x \pm \sqrt{140}$$

   $$y = 4x \pm 2\sqrt{35}$$

   So the tangent line equations are $y = 4x + 2\sqrt{35}$ or $y = 4x - 2\sqrt{35}$.

3. Answer:

   Given: ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and point $A(1, 3)$.

   First check whether point $A(1, 3)$ is on the ellipse:

   $$\frac{1^2}{9} + \frac{3^2}{4} = \frac{1}{9} + \frac{9}{4} = \frac{4 + 81}{36} = \frac{85}{36} > 1$$

   Since $\frac{85}{36} > 1$, point $A$ is outside the ellipse.

   Use the ellipse polar equation:

   $$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$

   Substitute $x_1 = 1$ and $y_1 = 3$:

   $$\frac{x \cdot 1}{9} + \frac{y \cdot 3}{4} = 1$$

   $$\frac{x}{9} + \frac{3y}{4} = 1$$

   $$4x + 27y = 36$$

   If you want to find the tangent points, substitute this equation into the ellipse equation. From $4x + 27y = 36$, we get $x = \frac{36 - 27y}{4}$.

   Substitute into the ellipse equation and solve to get two tangent points.

   So the tangent line equation is $4x + 27y = 36$.

4. Answer:

   Given: parabola $y^2 = 8x$ and point $A(2, 5)$.

   First check whether point $A(2, 5)$ is on the parabola:

   $$5^2 = 25 \neq 8 \cdot 2 = 16$$

   Point $A$ is outside the parabola.

   Use the parabola polar equation with $p = 2$:

   $$yy_1 = 4(x + x_1)$$

   Substitute $x_1 = 2$ and $y_1 = 5$:

   $$5y = 4(x + 2)$$

   $$5y = 4x + 8$$

   $$y = \frac{4x + 8}{5}$$

   Substitute into the parabola equation:

   $$\left(\frac{4x + 8}{5}\right)^2 = 8x$$

   $$\frac{(4x + 8)^2}{25} = 8x$$

   $$(4x + 8)^2 = 200x$$

   $$16x^2 + 64x + 64 = 200x$$

   $$16x^2 - 136x + 64 = 0$$

   $$2x^2 - 17x + 8 = 0$$

   Use the quadratic formula:

   $$x = \frac{17 \pm \sqrt{289 - 64}}{4} = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}$$

   So $x_1 = \frac{1}{2}$ and $x_2 = 8$.

   The tangent points are at $(\frac{1}{2}, 2)$ and $(8, 8)$.

   Tangent line equation through $(\frac{1}{2}, 2)$:

   $$y - 2 = \frac{2 - 5}{\frac{1}{2} - 2}(x - \frac{1}{2}) = \frac{-3}{-\frac{3}{2}}(x - \frac{1}{2}) = 2(x - \frac{1}{2})$$

   So $y = 2x + 1$.

   Tangent line equation through $(8, 8)$:

   $$y - 8 = \frac{8 - 5}{8 - 2}(x - 8) = \frac{3}{6}(x - 8) = \frac{1}{2}(x - 8)$$

   So $y = \frac{1}{2}x + 4$.

   In conclusion, the tangent line equations are $y = 2x + 1$ and $y = \frac{1}{2}x + 4$.