Topics

Tangents and normals to conics

Covers AS91573

What is being asked

A tangent touches the curve at one point and has the same gradient as the curve there.
A normal is perpendicular to the tangent at that point, so

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

Both are straight lines, so both answers have the form $y - y_1 = m(x - x_1)$ once you have the gradient and the point.

Three routes to the gradient

Route 1 — geometry (circles only). The tangent is perpendicular to the radius:
- Find the gradient of the radius from the centre to the point.
- Take the negative reciprocal.
- No calculus needed, and it is always the fastest route for a circle.
Route 2 — the discriminant. Substitute the line $y = mx + c$ $y = m x + c$ into the conic, and impose $\Delta = 0$ $Δ = 0$ for tangency.
- Use this when the point of contact is unknown — "find the tangents from an external point", or "for what $c$ is this line a tangent?"
Route 3 — implicit differentiation. Differentiate the conic equation term by term, treating $y$ $y$ as a function of $x$ $x$ .
- This belongs to AS91578, but is permitted and often quickest when the point of contact is known.
- Every result on this page is also obtainable by Routes 1 or 2, so the standard can be completed without it.

Route 1 in practice — the circle

Worked through — the tangent to $x^2 + y^2 = 25$ $x^{2} + y^{2} = 25$ at $(3, 4)$ $(3, 4)$ :
- Gradient of the radius from $(0,0)$ to $(3,4)$ : $m_r = \dfrac{4}{3}$
- Tangent gradient — the negative reciprocal: $m_t = -\dfrac{3}{4}$
- Equation: $y - 4 = -\dfrac{3}{4}(x - 3)$ , which tidies to $3x + 4y = 25$
- The normal is the radius itself, extended: $y = \dfrac{4}{3}x$
For a circle the normal always passes through the centre. That is a useful check and often the whole answer.

Route 2 in practice — the discriminant

Worked through — for what values of $c$ $c$ is $y = 2x + c$ $y = 2 x + c$ a tangent to $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ ?
- Substitute: $\dfrac{x^2}{9} + \dfrac{(2x+c)^2}{4} = 1$
- Multiply by 36 to clear fractions: $4x^2 + 9(2x+c)^2 = 36$
- Expand: $4x^2 + 9(4x^2 + 4cx + c^2) = 36$
- Collect: $40x^2 + 36cx + (9c^2 - 36) = 0$
- Set $\Delta = 0$ : $(36c)^2 - 4(40)(9c^2 - 36) = 0$
- $1296c^2 - 1440c^2 + 5760 = 0$
- $-144c^2 = -5760$ , so $c^2 = 40$ and $c = \pm 2\sqrt{10}$
Two answers, because there are two parallel tangents of any given gradient — one on each side of the ellipse.
The general condition for $y = mx + c$ to touch $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is

$c^2 = a^2m^2 + b^2$

Checking: $9(4) + 4 = 40$ ✓ — the same answer with far less algebra.

Route 3 in practice — implicit differentiation

Differentiate every term with respect to $x$ , and whenever you differentiate a $y$ term, multiply by $\dfrac{dy}{dx}$ .
Worked through — the tangent to $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ at $\left(4,\ \tfrac{9}{5}\right)$ $(4, \frac{9}{5})$ :
- Differentiate: $\dfrac{2x}{25} + \dfrac{2y}{9}\dfrac{dy}{dx} = 0$
- Rearrange: $\dfrac{dy}{dx} = -\dfrac{9x}{25y}$
- At the point: $\dfrac{dy}{dx} = -\dfrac{9(4)}{25(1.8)} = -\dfrac{36}{45} = -0.8$
- Tangent: $y - 1.8 = -0.8(x - 4)$ , so $y = -0.8x + 5$ , or $4x + 5y = 25$

The standard tangent formulas

Substituting $x^2 \to xx_1$ and $y^2 \to yy_1$ in the conic's equation gives the tangent at $(x_1, y_1)$ directly:

Conic	Tangent at $(x_1, y_1)$
$x^2 + y^2 = r^2$	$xx_1 + yy_1 = r^2$
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$	$\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$	$\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$
$y^2 = 4ax$	$yy_1 = 2a(x + x_1)$

Check the example above: $\dfrac{4x}{25} + \dfrac{(9/5)y}{9} = 1$ gives $\dfrac{4x}{25} + \dfrac{y}{5} = 1$ , and multiplying by 25 gives $4x + 5y = 25$ ✓
Use the formula to check, and show the working method in full. An internal wants the reasoning, not just a substitution into a memorised result.

Tangents at a parametric point

This is where the general results come from. For $y^2 = 4ax$ with $x = at^2$ , $y = 2at$ :

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$

Tangent at parameter $t$ :

$y - 2at = \frac{1}{t}\left(x - at^2\right) \quad \Rightarrow \quad ty = x + at^2$

Normal at parameter $t$ : gradient $-t$ , giving

$y + tx = 2at + at^3$

A general point labelled $t$ turns a statement about one point into a statement about every point, which is exactly what an Excellence proof needs.

Tips and Tricks

For a circle, use the perpendicular radius. It is faster than every other method and it is the property the standard is testing. The normal is then simply the line through the centre.

Two parallel tangents exist for any gradient on a closed conic, so a discriminant question of the form "for what $c$ …" should normally produce two answers. One answer usually means a sign was lost.

A negative reciprocal, not a negative. If the tangent gradient is $-\frac{3}{4}$ , the normal gradient is $+\frac{4}{3}$ — flip and change sign.

Keep exact values. $c = \pm2\sqrt{10}$ is a better answer than $\pm 6.32$ , and exact values make the final check clean.

Check that your point is actually on the curve before finding a tangent there. Substituting the given point into the conic takes one line and catches a misread coordinate.

Verify the tangent with the discriminant. Substituting your finished tangent back into the conic must give $\Delta = 0$ . It is the definitive check that the line touches rather than cuts.

Worked ExampleA tangent and normal on an ellipse

For the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ , find the equations of the tangent and the normal at the point $P(4,\ 1.8)$ , and find where the normal crosses the $x$ -axis.

Step 1 — Confirm the point lies on the curve

$\frac{4^2}{25} + \frac{1.8^2}{9} = \frac{16}{25} + \frac{3.24}{9} = 0.64 + 0.36 = 1 \ \checkmark$

$P$ is on the ellipse, so a tangent there exists.

Step 2 — Find the tangent gradient by implicit differentiation

Differentiate every term with respect to $x$ , multiplying by $\dfrac{dy}{dx}$ whenever a $y$ is differentiated:

$\frac{2x}{25} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0$

Make $\dfrac{dy}{dx}$ the subject:

$\frac{2y}{9} \cdot \frac{dy}{dx} = -\frac{2x}{25}$

$\frac{dy}{dx} = -\frac{2x}{25} \cdot \frac{9}{2y} = -\frac{9x}{25y}$

Step 3 — Evaluate at $P(4, 1.8)$

$\frac{dy}{dx} = -\frac{9(4)}{25(1.8)} = -\frac{36}{45} = -\frac{4}{5} = -0.8$

Sense-check: $P$ is in the first quadrant on an ellipse, where the curve is falling as $x$ increases — a negative gradient is right ✓

Step 4 — Write the tangent equation

$y - 1.8 = -0.8(x - 4)$

$y = -0.8x + 3.2 + 1.8$

$y = -0.8x + 5$

In tidy form, multiplying by 5:

$\boxed{4x + 5y = 25}$

Step 5 — Verify with the standard tangent formula

For an ellipse, the tangent at $(x_1, y_1)$ is $\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$ :

$\frac{4x}{25} + \frac{1.8y}{9} = 1$

$\frac{4x}{25} + \frac{y}{5} = 1$

Multiply by 25:

$4x + 5y = 25 \ \checkmark$

The two methods agree.

Step 6 — Confirm tangency with the discriminant

Substitute $y = \dfrac{25 - 4x}{5}$ into the ellipse:

$\frac{x^2}{25} + \frac{(25-4x)^2}{25 \times 9} = 1$

Multiply by 225:

$9x^2 + (25 - 4x)^2 = 225$

$9x^2 + 625 - 200x + 16x^2 = 225$

$25x^2 - 200x + 400 = 0$

Divide by 25:

$x^2 - 8x + 16 = 0 \quad \Rightarrow \quad (x - 4)^2 = 0$

Step 7 — Find the normal

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

$m_n = -\frac{1}{-0.8} = \frac{5}{4} = 1.25$

Equation through $P(4, 1.8)$ :

$y - 1.8 = 1.25(x - 4)$

$y = 1.25x - 5 + 1.8$

$\boxed{y = 1.25x - 3.2}$

Step 8 — Find where the normal crosses the $x$ -axis

Set $y = 0$ :

$0 = 1.25x - 3.2$

$x = \frac{3.2}{1.25} = 2.56$

Step 9 — Note a property worth observing

Unlike a circle, the normal to an ellipse does NOT pass through the centre. Here it crosses the axis at $2.56$ , not at the origin.

The general result is that the normal at $(x_1, y_1)$ meets the $x$ -axis at

$x = \frac{(a^2 - b^2)x_1}{a^2} = e^2 x_1$

Checking: $e^2 = \dfrac{a^2 - b^2}{a^2} = \dfrac{25 - 9}{25} = 0.64$ , and $0.64 \times 4 = 2.56$ ✓

Tangents and normals to conics

Covers AS91573

What is being asked

A tangent touches the curve at one point and has the same gradient as the curve there.
A normal is perpendicular to the tangent at that point, so

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

Both are straight lines, so both answers have the form $y - y_1 = m(x - x_1)$ once you have the gradient and the point.

Three routes to the gradient

Route 1 — geometry (circles only). The tangent is perpendicular to the radius:
- Find the gradient of the radius from the centre to the point.
- Take the negative reciprocal.
- No calculus needed, and it is always the fastest route for a circle.
Route 2 — the discriminant. Substitute the line $y = mx + c$ $y = m x + c$ into the conic, and impose $\Delta = 0$ $Δ = 0$ for tangency.
- Use this when the point of contact is unknown — "find the tangents from an external point", or "for what $c$ is this line a tangent?"
Route 3 — implicit differentiation. Differentiate the conic equation term by term, treating $y$ $y$ as a function of $x$ $x$ .
- This belongs to AS91578, but is permitted and often quickest when the point of contact is known.
- Every result on this page is also obtainable by Routes 1 or 2, so the standard can be completed without it.

Route 1 in practice — the circle

Worked through — the tangent to $x^2 + y^2 = 25$ $x^{2} + y^{2} = 25$ at $(3, 4)$ $(3, 4)$ :
- Gradient of the radius from $(0,0)$ to $(3,4)$ : $m_r = \dfrac{4}{3}$
- Tangent gradient — the negative reciprocal: $m_t = -\dfrac{3}{4}$
- Equation: $y - 4 = -\dfrac{3}{4}(x - 3)$ , which tidies to $3x + 4y = 25$
- The normal is the radius itself, extended: $y = \dfrac{4}{3}x$
For a circle the normal always passes through the centre. That is a useful check and often the whole answer.

Route 2 in practice — the discriminant

Worked through — for what values of $c$ $c$ is $y = 2x + c$ $y = 2 x + c$ a tangent to $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ ?
- Substitute: $\dfrac{x^2}{9} + \dfrac{(2x+c)^2}{4} = 1$
- Multiply by 36 to clear fractions: $4x^2 + 9(2x+c)^2 = 36$
- Expand: $4x^2 + 9(4x^2 + 4cx + c^2) = 36$
- Collect: $40x^2 + 36cx + (9c^2 - 36) = 0$
- Set $\Delta = 0$ : $(36c)^2 - 4(40)(9c^2 - 36) = 0$
- $1296c^2 - 1440c^2 + 5760 = 0$
- $-144c^2 = -5760$ , so $c^2 = 40$ and $c = \pm 2\sqrt{10}$
Two answers, because there are two parallel tangents of any given gradient — one on each side of the ellipse.
The general condition for $y = mx + c$ to touch $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is

$c^2 = a^2m^2 + b^2$

Checking: $9(4) + 4 = 40$ ✓ — the same answer with far less algebra.

Route 3 in practice — implicit differentiation

Differentiate every term with respect to $x$ , and whenever you differentiate a $y$ term, multiply by $\dfrac{dy}{dx}$ .
Worked through — the tangent to $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ at $\left(4,\ \tfrac{9}{5}\right)$ $(4, \frac{9}{5})$ :
- Differentiate: $\dfrac{2x}{25} + \dfrac{2y}{9}\dfrac{dy}{dx} = 0$
- Rearrange: $\dfrac{dy}{dx} = -\dfrac{9x}{25y}$
- At the point: $\dfrac{dy}{dx} = -\dfrac{9(4)}{25(1.8)} = -\dfrac{36}{45} = -0.8$
- Tangent: $y - 1.8 = -0.8(x - 4)$ , so $y = -0.8x + 5$ , or $4x + 5y = 25$

The standard tangent formulas

Substituting $x^2 \to xx_1$ and $y^2 \to yy_1$ in the conic's equation gives the tangent at $(x_1, y_1)$ directly:

Conic	Tangent at $(x_1, y_1)$
$x^2 + y^2 = r^2$	$xx_1 + yy_1 = r^2$
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$	$\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$	$\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$
$y^2 = 4ax$	$yy_1 = 2a(x + x_1)$

Check the example above: $\dfrac{4x}{25} + \dfrac{(9/5)y}{9} = 1$ gives $\dfrac{4x}{25} + \dfrac{y}{5} = 1$ , and multiplying by 25 gives $4x + 5y = 25$ ✓
Use the formula to check, and show the working method in full. An internal wants the reasoning, not just a substitution into a memorised result.

Tangents at a parametric point

This is where the general results come from. For $y^2 = 4ax$ with $x = at^2$ , $y = 2at$ :

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$

Tangent at parameter $t$ :

$y - 2at = \frac{1}{t}\left(x - at^2\right) \quad \Rightarrow \quad ty = x + at^2$

Normal at parameter $t$ : gradient $-t$ , giving

$y + tx = 2at + at^3$

A general point labelled $t$ turns a statement about one point into a statement about every point, which is exactly what an Excellence proof needs.

Tips and Tricks

For a circle, use the perpendicular radius. It is faster than every other method and it is the property the standard is testing. The normal is then simply the line through the centre.

A negative reciprocal, not a negative. If the tangent gradient is $-\frac{3}{4}$ , the normal gradient is $+\frac{4}{3}$ — flip and change sign.

Keep exact values. $c = \pm2\sqrt{10}$ is a better answer than $\pm 6.32$ , and exact values make the final check clean.

Check that your point is actually on the curve before finding a tangent there. Substituting the given point into the conic takes one line and catches a misread coordinate.

Verify the tangent with the discriminant. Substituting your finished tangent back into the conic must give $\Delta = 0$ . It is the definitive check that the line touches rather than cuts.

Worked ExampleA tangent and normal on an ellipse

For the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ , find the equations of the tangent and the normal at the point $P(4,\ 1.8)$ , and find where the normal crosses the $x$ -axis.

Step 1 — Confirm the point lies on the curve

$\frac{4^2}{25} + \frac{1.8^2}{9} = \frac{16}{25} + \frac{3.24}{9} = 0.64 + 0.36 = 1 \ \checkmark$

$P$ is on the ellipse, so a tangent there exists.

Step 2 — Find the tangent gradient by implicit differentiation

Differentiate every term with respect to $x$ , multiplying by $\dfrac{dy}{dx}$ whenever a $y$ is differentiated:

$\frac{2x}{25} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0$

Make $\dfrac{dy}{dx}$ the subject:

$\frac{2y}{9} \cdot \frac{dy}{dx} = -\frac{2x}{25}$

$\frac{dy}{dx} = -\frac{2x}{25} \cdot \frac{9}{2y} = -\frac{9x}{25y}$

Step 3 — Evaluate at $P(4, 1.8)$

$\frac{dy}{dx} = -\frac{9(4)}{25(1.8)} = -\frac{36}{45} = -\frac{4}{5} = -0.8$

Sense-check: $P$ is in the first quadrant on an ellipse, where the curve is falling as $x$ increases — a negative gradient is right ✓

Step 4 — Write the tangent equation

$y - 1.8 = -0.8(x - 4)$

$y = -0.8x + 3.2 + 1.8$

$y = -0.8x + 5$

In tidy form, multiplying by 5:

$\boxed{4x + 5y = 25}$

Step 5 — Verify with the standard tangent formula

For an ellipse, the tangent at $(x_1, y_1)$ is $\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$ :

$\frac{4x}{25} + \frac{1.8y}{9} = 1$

$\frac{4x}{25} + \frac{y}{5} = 1$

Multiply by 25:

$4x + 5y = 25 \ \checkmark$

The two methods agree.

Step 6 — Confirm tangency with the discriminant

Substitute $y = \dfrac{25 - 4x}{5}$ into the ellipse:

$\frac{x^2}{25} + \frac{(25-4x)^2}{25 \times 9} = 1$

Multiply by 225:

$9x^2 + (25 - 4x)^2 = 225$

$9x^2 + 625 - 200x + 16x^2 = 225$

$25x^2 - 200x + 400 = 0$

Divide by 25:

$x^2 - 8x + 16 = 0 \quad \Rightarrow \quad (x - 4)^2 = 0$

Step 7 — Find the normal

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

$m_n = -\frac{1}{-0.8} = \frac{5}{4} = 1.25$

Equation through $P(4, 1.8)$ :

$y - 1.8 = 1.25(x - 4)$

$y = 1.25x - 5 + 1.8$

$\boxed{y = 1.25x - 3.2}$

Step 8 — Find where the normal crosses the $x$ -axis

Set $y = 0$ :

$0 = 1.25x - 3.2$

$x = \frac{3.2}{1.25} = 2.56$

Step 9 — Note a property worth observing

Unlike a circle, the normal to an ellipse does NOT pass through the centre. Here it crosses the axis at $2.56$ , not at the origin.

The general result is that the normal at $(x_1, y_1)$ meets the $x$ -axis at

$x = \frac{(a^2 - b^2)x_1}{a^2} = e^2 x_1$

Checking: $e^2 = \dfrac{a^2 - b^2}{a^2} = \dfrac{25 - 9}{25} = 0.64$ , and $0.64 \times 4 = 2.56$ ✓

Tangents and normals to conics

What is being asked

Three routes to the gradient

Route 1 in practice — the circle

Route 2 in practice — the discriminant

Route 3 in practice — implicit differentiation

The standard tangent formulas

Tangents at a parametric point

Step 1 — Confirm the point lies on the curve

Step 2 — Find the tangent gradient by implicit differentiation

Step 3 — Evaluate at P(4,1.8)P(4, 1.8)P(4,1.8)

Step 4 — Write the tangent equation

Step 5 — Verify with the standard tangent formula

Step 6 — Confirm tangency with the discriminant

Step 7 — Find the normal

Step 8 — Find where the normal crosses the xxx-axis

Step 9 — Note a property worth observing

Tangents and normals to conics

What is being asked

Three routes to the gradient

Route 1 in practice — the circle

Route 2 in practice — the discriminant

Route 3 in practice — implicit differentiation

The standard tangent formulas

Tangents at a parametric point

Step 1 — Confirm the point lies on the curve

Step 2 — Find the tangent gradient by implicit differentiation

Step 3 — Evaluate at P(4,1.8)P(4, 1.8)P(4,1.8)

Step 4 — Write the tangent equation

Step 5 — Verify with the standard tangent formula

Step 6 — Confirm tangency with the discriminant

Step 7 — Find the normal

Step 8 — Find where the normal crosses the xxx-axis

Step 9 — Note a property worth observing

Step 3 — Evaluate at $P(4, 1.8)$

Step 8 — Find where the normal crosses the $x$ -axis

Step 3 — Evaluate at $P(4, 1.8)$

Step 8 — Find where the normal crosses the $x$ -axis