Tangent Lines to Conic Sections

Repeated Roots Identify Tangency

A ruler placed against a smooth curve can meet it at just one point without crossing it nearby. The line along the ruler models a tangent line at that point.

After a line equation is substituted into the equation of a parabola, ellipse, or hyperbola, a tangent produces a repeated root. A secant produces two distinct roots. On the parabola $y = x^2$ , the repeated root identifies the single point where the tangent touches the curve.

The given information determines the method: a known tangent point gives a point form, a known slope gives a line with an unknown intercept, and an external point gives a family of lines through that point.

Point on the Curve

When the point of tangency is known, implicit differentiation gives the slope there. Combining that slope with the point-slope equation produces a direct point form.

Point Form from Implicit Differentiation

For a point $T(x_1, y_1)$ on a conic, the resulting point form replaces each squared coordinate by a product with the corresponding tangent-point coordinate. Linear terms are shared symmetrically between the variable point and the tangent point. This compact form comes from the derivative calculation. Before using it, check that $T$ lies on the curve.

For the parabola $y^2 = 4px$ at the tangent point $T(x_1, y_1)$ , the tangent line equation is:

yy_1 = 2p(x + x_1)

Apply the formula to the parabola $y^2 = 8x$ at $(2, 4)$ .

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

y \cdot 4 = 2 \cdot 2 \cdot (x + 2)

The point form is valid only after the stated point has been verified on the conic. It packages the derivative calculation into a reusable algebraic substitution.

Given Slope

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 tangents to $4x^2 - 9y^2 = 36$ that are perpendicular to $x + 4y + 10 = 0$ .

First determine the required tangent slope. The line $x + 4y + 10 = 0$ has slope $m_g = -\frac{1}{4}$ , so every perpendicular tangent must have slope $m_s = 4$ .

A line with slope $m$ has the form $y = mx + c$ . Substitute it into the hyperbola equation:

4x^2 - 9(4x + c)^2 = 36

For $y=4x+c$ to be tangent to the hyperbola, the substituted quadratic equation must have a repeated root. Its discriminant must therefore be zero:

\begin{aligned} D &= (-72c)^2 - 4(-140)(-9c^2-36) \\ &= 0 \end{aligned}

The two tangent lines are $y = 4x + 2\sqrt{35}$ and $y = 4x - 2\sqrt{35}$ .

Tangents Through an External Point

Depending on the conic and the point's location, zero, one, or two real tangents can pass through a given point. In the example below, there are two. Drawing tangents from a point outside a circle is a familiar special case.

For the parabola $y^2 = 8x$ , two different tangent lines pass through $A(2, 5)$ .

For an external point, the polar equation gives the chord of contact, the line joining the two tangent points. It is not itself either tangent. For $y^2 = 8x$ and $A(2,5)$ , the polar is:

yy_1 = 4(x + x_1)

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

5y = 4(x + 2)

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

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

Using the quadratic formula, we get $x_1 = \frac{1}{2}$ and $x_2 = 8$ . The tangent points are $(\frac{1}{2}, 2)$ and $(8, 8)$ . Joining each one to $A(2,5)$ gives the two tangents $y = 2x + 1$ and $y = \frac{1}{2}x + 4$ .

Tangent Line Equations for Each Conic Section

The table summarizes tangent-line equations written in point form.

Conic Section	Tangent Line Equation
$y^2 = 4px$	$yy_1 = 2p(x + x_1)$
$x^2 = 4py$	$xx_1 = 2p(y + y_1)$
$(x - h)^2 = 4p(y - k)$	$(x - h)(x_1 - h) = 2p((y - k) + (y_1 - k))$
$(y - k)^2 = 4p(x - h)$	$(y - k)(y_1 - k) = 2p((x - h) + (x_1 - h))$
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$	$\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$
$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$	$\frac{yy_1}{a^2} - \frac{xx_1}{b^2} = 1$
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$	$\frac{(x-h)(x_1-h)}{a^2} + \frac{(y-k)(y_1-k)}{b^2} = 1$
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$	$\frac{(x-h)(x_1-h)}{a^2} - \frac{(y-k)(y_1-k)}{b^2} = 1$
$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$	$\frac{(y-k)(y_1-k)}{a^2} - \frac{(x-h)(x_1-h)}{b^2} = 1$

These formulas are the derivative-based point forms for common standard conics. Always verify first that the stated tangent point lies on the conic.

Exercises

Each problem gives a conic section and a point on it, and asks for the tangent line at that point. Write the conic in standard form before applying the tangent formula.

Find the tangent line equation of parabola $y^2 = 12x$ at point $(3, 6)$ .
Find the equations of the tangents to $4x^2 - 9y^2 = 36$ that are perpendicular to $x + 4y + 10 = 0$ .
Find the equations of the tangents to $\frac{x^2}{9} + \frac{y^2}{4} = 1$ that pass through $A(1, 3)$ .
Find the equations of the tangents to $y^2 = 8x$ that pass through $A(2, 5)$ .

Worked Solutions

Solution:

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:
$yy_1 = 2p(x + x_1)$
So the tangent line equation is $y = x + 3$ .
Solution:

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 $m = 4$ :
$y = 4x \pm \sqrt{9 \cdot 16 - 4}$
So the tangent line equations are $y = 4x + 2\sqrt{35}$ or $y = 4x - 2\sqrt{35}$ .
Solution:

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:

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

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

Write a line through $A(1,3)$ with slope $m$ :

$y-3=m(x-1)$

Substituting this line into the ellipse and requiring a zero discriminant gives:
$4x^2 + 9(mx+3-m)^2 = 36$
Insert each slope into the point-slope equation:
$\begin{aligned} &y-3=-\frac{5}{4}(x-1) \\ &\Longrightarrow \\ &5x+4y-17=0 \end{aligned}$
Thus the two tangent lines are $5x+4y-17=0$ and $x-2y+5=0$ .
Solution:

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 $y_1 = 5$ :
$5y = 4(x + 2)$
Substitute into the parabola equation:
$\left(\frac{4x + 8}{5}\right)^2 = 8x$
Use the quadratic formula:

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

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 $(\frac{1}{2}, 2)$ :

$\begin{aligned} y - 2 &= \frac{2 - 5}{\frac{1}{2} - 2}(x - \frac{1}{2}) \\ &= \frac{-3}{-\frac{3}{2}}(x - \frac{1}{2}) \\ &= 2(x - \frac{1}{2}) \end{aligned}$

So $y = 2x + 1$ .

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

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

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

The tangent line equations are $y = 2x + 1$ and $y = \frac{1}{2}x + 4$ .