# Nakafa Framework: LLM

URL: /en/subject/high-school/12/mathematics/analytic-geometry/tangent-line-to-conic-sections
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---

export const metadata = {
  title: "Tangent Line to Conic Sections",
  description: "Master tangent lines for parabolas, ellipses, and hyperbolas using the division principle. Find equations for points on curves, given slopes, and external points.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/26/2025",
  subject: "Analytic Geometry",
};

## 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 <InlineMath math="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 <InlineMath math="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 <InlineMath math="y^2 = 4px" /> and tangent point <InlineMath math="T(x_1, y_1)" />, then the tangent line equation becomes:

<BlockMath math="yy_1 = 2p(x + x_1)" />

Let's take an example of parabola <InlineMath math="y^2 = 8x" /> at point <InlineMath math="(2, 4)" />.

From parabola <InlineMath math="y^2 = 8x" />, we know <InlineMath math="4p = 8" /> so <InlineMath math="p = 2" />. The tangent line at point <InlineMath math="(2, 4)" /> is:

<div className="flex flex-col gap-4">
  <BlockMath math="y \cdot 4 = 2 \cdot 2 \cdot (x + 2)" />
  <BlockMath math="4y = 4(x + 2)" />
  <BlockMath math="y = x + 2" />
</div>

> 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 <InlineMath math="m" /> into the conic section equation.

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

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

The tangent line equation of a hyperbola with gradient <InlineMath math="m" /> is <InlineMath math="y = mx + c" />. We substitute it into the hyperbola equation:

<div className="flex flex-col gap-4">
  <BlockMath math="4x^2 - 9(4x + c)^2 = 36" />
  <BlockMath math="4x^2 - 9(16x^2 + 8cx + c^2) = 36" />
  <BlockMath math="-140x^2 - 72cx - 9c^2 = 36" />
</div>

For a tangent line, the discriminant must be zero:

<BlockMath math="D = (-72c)^2 - 4(-140)(-9c^2) = 0" />

So we get <InlineMath math="c = 0" /> and the tangent line equation is <InlineMath math="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 <InlineMath math="y^2 = 8x" /> with point <InlineMath math="A(2, 5)" />. From point <InlineMath math="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 <InlineMath math="y^2 = 8x" /> with point <InlineMath math="A(2,5)" />, the polar equation is:

<BlockMath math="yy_1 = 4(x + x_1)" />

Substitute <InlineMath math="x_1 = 2" /> and <InlineMath math="y_1 = 5" />:

<div className="flex flex-col gap-4">
  <BlockMath math="5y = 4(x + 2)" />
  <BlockMath math="5y = 4x + 8" />
  <BlockMath math="y = \frac{4x + 8}{5}" />
</div>

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

<div className="flex flex-col gap-4">
  <BlockMath math="\left(\frac{4x + 8}{5}\right)^2 = 8x" />
  <BlockMath math="\frac{16x^2 + 64x + 64}{25} = 8x" />
  <BlockMath math="16x^2 + 64x + 64 = 200x" />
  <BlockMath math="16x^2 - 136x + 64 = 0" />
  <BlockMath math="2x^2 - 17x + 8 = 0" />
</div>

Using the quadratic formula, we get <InlineMath math="x_1 = \frac{1}{2}" /> and <InlineMath math="x_2 = 8" />. So the tangent points are at <InlineMath math="(\frac{1}{2}, 2)" /> and <InlineMath math="(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!

| Conic Section | Tangent Line Equation |
|---|---|
| <InlineMath math="y^2 = 4px" /> | <InlineMath math="yy_1 = 2p(x + x_1)" /> |
| <InlineMath math="x^2 = 4py" /> | <InlineMath math="xx_1 = 2p(y + y_1)" /> |
| <InlineMath math="(x - m)^2 = 4p(y - n)" /> | <InlineMath math="(x - m)(x_1 - m) = 2p(y - n)(y_1 - n)" /> |
| <InlineMath math="(y - m)^2 = 4p(x - n)" /> | <InlineMath math="(y - m)(y_1 - m) = 2p(x - n)(x_1 - n)" /> |
| <InlineMath math="\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1" /> | <InlineMath math="\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1" /> |
| <InlineMath math="\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1" /> | <InlineMath math="\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1" /> |
| <InlineMath math="\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1" /> | <InlineMath math="\frac{yy_1}{a^2} - \frac{xx_1}{b^2} = 1" /> |
| <InlineMath math="\frac{(x-m)^2}{a^2} + \frac{(y-n)^2}{b^2} = 1" /> | <InlineMath math="\frac{(x-m)(x_1-m)}{a^2} + \frac{(y-n)(y_1-n)}{b^2} = 1" /> |
| <InlineMath math="\frac{(x-m)^2}{a^2} - \frac{(y-n)^2}{b^2} = 1" /> | <InlineMath math="\frac{(x-m)(x_1-m)}{a^2} - \frac{(y-n)(y_1-n)}{b^2} = 1" /> |
| <InlineMath math="\frac{(y-n)^2}{a^2} - \frac{(x-m)^2}{b^2} = 1" /> | <InlineMath math="\frac{(y-n)(y_1-n)}{a^2} - \frac{(x-m)(x_1-m)}{b^2} = 1" /> |

> 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 <InlineMath math="y^2 = 12x" /> at point <InlineMath math="(3, 6)" />.

2. There's a hyperbola <InlineMath math="4x^2 - 9y^2 = 36" />. Find the tangent line equation that is perpendicular to line <InlineMath math="x + 4y + 10 = 0" />.

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

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

### Answer Key

1. **Answer**:

   Given: parabola <InlineMath math="y^2 = 12x" /> with tangent point <InlineMath math="(3, 6)" />.
   
   From equation <InlineMath math="y^2 = 12x" />, we get <InlineMath math="4p = 12" /> so <InlineMath math="p = 3" />.
   
   Use the parabola tangent line formula:
   
   <div className="flex flex-col gap-4">
     <BlockMath math="yy_1 = 2p(x + x_1)" />
     <BlockMath math="y \cdot 6 = 2 \cdot 3 \cdot (x + 3)" />
     <BlockMath math="6y = 6(x + 3)" />
     <BlockMath math="6y = 6x + 18" />
     <BlockMath math="y = x + 3" />
   </div>
   
   So the tangent line equation is <InlineMath math="y = x + 3" />.

2. **Answer**:

   Given: hyperbola <InlineMath math="4x^2 - 9y^2 = 36" /> or <InlineMath math="\frac{x^2}{9} - \frac{y^2}{4} = 1" />.
   
   Line <InlineMath math="x + 4y + 10 = 0" /> has gradient <InlineMath math="m = -\frac{1}{4}" />.
   
   Since the tangent line is perpendicular to that line, then the tangent line gradient is:
   
   <BlockMath math="m_s = -\frac{1}{m} = -\frac{1}{-\frac{1}{4}} = 4" />
   
   Hyperbola tangent line formula with gradient <InlineMath math="m" />:
   
   <BlockMath math="y = mx \pm \sqrt{a^2m^2 - b^2}" />
   
   With <InlineMath math="a^2 = 9" />, <InlineMath math="b^2 = 4" />, and <InlineMath math="m = 4" />:
   
   <div className="flex flex-col gap-4">
     <BlockMath math="y = 4x \pm \sqrt{9 \cdot 16 - 4}" />
     <BlockMath math="y = 4x \pm \sqrt{144 - 4}" />
     <BlockMath math="y = 4x \pm \sqrt{140}" />
     <BlockMath math="y = 4x \pm 2\sqrt{35}" />
   </div>
   
   So the tangent line equations are <InlineMath math="y = 4x + 2\sqrt{35}" /> or <InlineMath math="y = 4x - 2\sqrt{35}" />.

3. **Answer**:

   Given: ellipse <InlineMath math="\frac{x^2}{9} + \frac{y^2}{4} = 1" /> and point <InlineMath math="A(1, 3)" />.
   
   First check whether point <InlineMath math="A(1, 3)" /> is on the ellipse:
   
   <BlockMath math="\frac{1^2}{9} + \frac{3^2}{4} = \frac{1}{9} + \frac{9}{4} = \frac{4 + 81}{36} = \frac{85}{36} > 1" />
   
   Since <InlineMath math="\frac{85}{36} > 1" />, point <InlineMath math="A" /> is outside the ellipse.
   
   Use the ellipse polar equation:
   
   <BlockMath math="\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1" />
   
   Substitute <InlineMath math="x_1 = 1" /> and <InlineMath math="y_1 = 3" />:
   
   <div className="flex flex-col gap-4">
     <BlockMath math="\frac{x \cdot 1}{9} + \frac{y \cdot 3}{4} = 1" />
     <BlockMath math="\frac{x}{9} + \frac{3y}{4} = 1" />
     <BlockMath math="4x + 27y = 36" />
   </div>
   
   If you want to find the tangent points, substitute this equation into the ellipse equation. From <InlineMath math="4x + 27y = 36" />, we get <InlineMath math="x = \frac{36 - 27y}{4}" />.

   Substitute into the ellipse equation and solve to get two tangent points.
   
   So the tangent line equation is <InlineMath math="4x + 27y = 36" />.

4. **Answer**:

   Given: parabola <InlineMath math="y^2 = 8x" /> and point <InlineMath math="A(2, 5)" />.
   
   First check whether point <InlineMath math="A(2, 5)" /> is on the parabola:
   
   <BlockMath math="5^2 = 25 \neq 8 \cdot 2 = 16" />
   
   Point <InlineMath math="A" /> is outside the parabola.
   
   Use the parabola polar equation with <InlineMath math="p = 2" />:
   
   <BlockMath math="yy_1 = 4(x + x_1)" />
   
   Substitute <InlineMath math="x_1 = 2" /> and <InlineMath math="y_1 = 5" />:
   
   <div className="flex flex-col gap-4">
     <BlockMath math="5y = 4(x + 2)" />
     <BlockMath math="5y = 4x + 8" />
     <BlockMath math="y = \frac{4x + 8}{5}" />
   </div>
   
   Substitute into the parabola equation:
   
   <div className="flex flex-col gap-4">
     <BlockMath math="\left(\frac{4x + 8}{5}\right)^2 = 8x" />
     <BlockMath math="\frac{(4x + 8)^2}{25} = 8x" />
     <BlockMath math="(4x + 8)^2 = 200x" />
     <BlockMath math="16x^2 + 64x + 64 = 200x" />
     <BlockMath math="16x^2 - 136x + 64 = 0" />
     <BlockMath math="2x^2 - 17x + 8 = 0" />
   </div>
   
   Use the quadratic formula:
   
   <BlockMath math="x = \frac{17 \pm \sqrt{289 - 64}}{4} = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}" />
   
   So <InlineMath math="x_1 = \frac{1}{2}" /> and <InlineMath math="x_2 = 8" />.
   
   The tangent points are at <InlineMath math="(\frac{1}{2}, 2)" /> and <InlineMath math="(8, 8)" />.
   
   Tangent line equation through <InlineMath math="(\frac{1}{2}, 2)" />:
   
   <BlockMath math="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 <InlineMath math="y = 2x + 1" />.
   
   Tangent line equation through <InlineMath math="(8, 8)" />:
   
   <BlockMath math="y - 8 = \frac{8 - 5}{8 - 2}(x - 8) = \frac{3}{6}(x - 8) = \frac{1}{2}(x - 8)" />
   
   So <InlineMath math="y = \frac{1}{2}x + 4" />.
   
   In conclusion, the tangent line equations are <InlineMath math="y = 2x + 1" /> and <InlineMath math="y = \frac{1}{2}x + 4" />.