FP1 4.3 - Tangents and normals to conics | Edexcel A-Level Further Mathematics

Use algebraic differentiation to find tangents and normals to the four standard conics. You will handle vertical tangents, derive the conditions for a line to touch a conic and find tangents through an external point.

A tangent direction and its perpendicular

Two lines through $(5,0)$ touch $x^2/16+y^2/9=1$ , and they are perpendicular. How can you find them when $(5,0)$ is not on the ellipse? First distinguish a tangent at a known curve point from a tangent through a point elsewhere.

You already know

From A-level Mathematics, a line of gradient $m$ through $(x_1,y_1)$ has equation $y-y_1=m(x-x_1)$ . A normal is perpendicular to the tangent: finite non-zero gradients have product $-1$ . Recall implicit differentiation and the parametric conic coordinates from FP1 4.1.

For an implicit equation, differentiate both coordinates as functions of $x$ ; a term $y^2$ becomes $2y\,\mathrm{d}y/\mathrm{d}x$ . Evaluate the gradient only at a point known to lie on the curve.

If the tangent is horizontal, its normal is vertical. If the tangent is vertical, its normal is horizontal. The phrase “negative reciprocal” does not give a numerical gradient in those two cases; use the directions themselves.

Key idea

At a given curve point, differentiate and use a point-line equation. Through an external point, the point is not a contact point: impose tangency as an additional condition.

Check the vertex before dividing by a coordinate that may be zero.

Parabola tangents and normals

Differentiate $y^2=4ax$ :

2y\frac{\mathrm{d}y}{\mathrm{d}x}=4a \quad\Longrightarrow\quad \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2a}{y}=\frac1t

at $(at^2,2at)$ , provided $t\ne0$ . The normal gradient is then $-t$ .

Lines at a parabola parameter

At $(at^2,2at)$ on $y^2=4ax$ , $a>0$ ,

\text{tangent:}\quad ty=x+at^2

\text{normal:}\quad y=-tx+2at+at^3

For $t\ne0$ , these follow by inserting gradients $1/t,-t$ into point-line equations. At $t=0$ , they correctly give the vertex tangent $x=0$ and normal $y=0$ .

Not in the formula booklet: learn it.

The tangent equation comes from $y-2at=(x-at^2)/t$ , multiplied by $t$ . Multiplication gives a useful formula that remains valid at the vertex, though its finite-gradient derivation required a separate vertex check.

[DIAGRAM: asset_name: A tangent and normal to a parabola; asset_slug: 9fm0-fp172-parabola-lines; caption: The tangent and normal pass through the same curve point and are perpendicular; description: An equal-scale plot shows y squared equals 8x, the point P at (2,4), its tangent y equals x plus 2 in blue and its normal y equals minus x plus 6 in sage. A right-angle marker at P separates the two line directions. A side panel gives the implicit derivative 4 over y, tangent gradient 1 and normal gradient minus 1. The vertex is also marked, with a note that its tangent is x equals zero and its normal y equals zero.]
The tangent and normal pass through the same curve point and are perpendicular

Worked example 1: two lines at a positive parameter

Question. Find the tangent and normal to $y^2=8x$ at $(2,4)$ .

Plan. Differentiate, evaluate the gradient, then use the point for both lines.

\begin{aligned} 2y\,y'&=8\quad\Longrightarrow\quad y'=\frac4y && \text{implicit differentiation}\\ m_T&=1,\qquad m_N=-1 && y=4\\ T:\ y-4&=x-2,\qquad N:\ y-4=-(x-2) && \text{point-line equations} \end{aligned}

Why this step? The normal changes the direction, not the point: it must pass through $(2,4)$ too.

Answer. Tangent $y=x+2$ ; normal $y=-x+6$ .

Check. Both contain $(2,4)$ , and their gradients multiply to $-1$ .

Change only the sign of the point's vertical coordinate, reflecting it in the horizontal axis.

Worked example 2: the reflected point

Question. Find the tangent and normal to $y^2=8x$ at $(2,-4)$ .

Plan. Use the same derivative with the signed value of $y$ .

\begin{aligned} m_T&=\frac4{-4}=-1,\qquad m_N=1 && \text{gradient signs}\\ T:\ y+4&=-(x-2),\qquad N:\ y+4=x-2 && \text{through the reflected point} \end{aligned}

Why this step? The sign of $y$ survives in $4/y$ ; the two reflected tangents have opposite gradients.

Answer. Tangent $y=-x-2$ ; normal $y=x-6$ .

Check. With $a=2,t=-1$ , the parameter formulas give the same two equations.

For a question saying “use calculus”, show the differentiation even when you recognise a standard line formula.

Watch out

Wrong: divide by $t=0$ to obtain the vertex gradient. Right: use the vertical tangent direction at the vertex and write its equation directly.

Move to the point with parameter $t=2$ .

Ellipse tangents and normals

Differentiating $x^2/a^2+y^2/b^2=1$ gives

\frac{2x}{a^2}+\frac{2y}{b^2}y'=0 \quad\Longrightarrow\quad y'=-\frac{b^2x}{a^2y}

where $y\ne0$ . Substitution into a point-line equation and use of the point's own curve equation yield the tangent below.

Lines at an ellipse point

At $(x_1,y_1)$ on $x^2/a^2+y^2/b^2=1$ ,

\text{tangent:}\quad \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1

\text{normal:}\quad b^2x_1(y-y_1)=a^2y_1(x-x_1)

The undivided normal equation includes the axis-intercept cases. At $(\pm a,0)$ the tangent is vertical and the normal horizontal; at $(0,\pm b)$ the directions reverse.

Not in the formula booklet: learn it.

For example, multiplying the tangent point-line equation by $y_1/b^2$ gives

\frac{yy_1-y_1^2}{b^2} =-\frac{xx_1-x_1^2}{a^2}

and the curve identity $x_1^2/a^2+y_1^2/b^2=1$ supplies its right-hand constant.

Worked example 3: an exact non-axis point

Question. Find the tangent and normal to $x^2/25+y^2/9=1$ at $(3,12/5)$ .

Plan. Evaluate the implicit derivative and form both lines.

\begin{aligned} m_T&=-\frac{9(3)}{25(12/5)}=-\frac9{20} && \text{differentiate and substitute}\\ T:\ \frac{3x}{25}+\frac{4y}{15}&=1 \quad\Longrightarrow\quad 9x+20y=75 && \text{tangent equation}\\ N:\ y-\frac{12}5&=\frac{20}9(x-3) && \text{negative reciprocal} \end{aligned}

Why this step? The normal gradient is $20/9$ , not $9/20$ : both the sign and reciprocal change.

Answer. Tangent $9x+20y=75$ ; normal $y-12/5=(20/9)(x-3)$ .

Check. The point satisfies both lines and $9/25+(144/25)/9=1$ .

Reflect the point vertically, changing only $y_1$ 's sign.

Worked example 4: the reflected ellipse point

Question. Find both lines at $(3,-12/5)$ on the same ellipse.

Plan. Retain the negative vertical coordinate in the derivative and tangent formula.

\begin{aligned} m_T&=\frac9{20},\qquad m_N=-\frac{20}9 && \text{changed sign}\\ T:\ 9x-20y&=75 && \text{tangent}\\ N:\ y+\frac{12}5&=-\frac{20}9(x-3) && \text{normal} \end{aligned}

Why this step? Reflection changes both line gradients; it does not merely change the sign of their constants.

Answer. The tangent and normal are as displayed.

Check. Substituting $(3,-12/5)$ into the tangent gives $27+48=75$ , and the normal's two sides are both zero.

The point formula can be used only when $(x_1,y_1)$ is on the ellipse. An external point cannot be inserted as though it were a contact point.

Watch out

Wrong: use the ellipse's tangent gradient for its normal too. Right: take the perpendicular direction, checking horizontal or vertical cases separately.

Use a new exact point on the same ellipse.

Hyperbola lines and the changed sign

For $x^2/a^2-y^2/b^2=1$ , differentiation gives

\frac{2x}{a^2}-\frac{2y}{b^2}y'=0 \quad\Longrightarrow\quad y'=\frac{b^2x}{a^2y}

The minus sign in the curve changes the tangent gradient's sign compared with the ellipse calculation.

Lines at a hyperbola point

At $(x_1,y_1)$ on $x^2/a^2-y^2/b^2=1$ ,

\text{tangent:}\quad \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1

\text{normal:}\quad b^2x_1(y-y_1)=-a^2y_1(x-x_1)

At the vertices $(\pm a,0)$ , the tangents are $x=\pm a$ and the normals $y=0$ .

Not in the formula booklet: learn it.

The tangent formula follows from the derivative and the identity satisfied by the contact point, just as for the ellipse. Keeping the original curve's sign through the rearrangement is essential.

Worked example 5: the right branch

Question. Find the tangent and normal to $x^2/9-y^2/4=1$ at $(5,8/3)$ .

Plan. Differentiate the hyperbola, evaluate its gradient and use the point.

\begin{aligned} m_T&=\frac{4(5)}{9(8/3)}=\frac56,\qquad m_N=-\frac65 && \text{implicit derivative}\\ T:\ \frac{5x}9-\frac{2y}3&=1 \quad\Longrightarrow\quad 5x-6y=9 && \text{tangent}\\ N:\ y-\frac83&=-\frac65(x-5) && \text{normal} \end{aligned}

Why this step? Both coordinates are positive, so the right upper branch has positive tangent gradient; the ellipse's extra minus sign would be wrong here.

Answer. Tangent $5x-6y=9$ ; normal $y-8/3=-(6/5)(x-5)$ .

Check. The point gives $25-16=9$ in the tangent and $25/9-16/9=1$ in the hyperbola.

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FP1 4.3 - Tangents and normals to conics

A tangent direction and its perpendicular

Parabola tangents and normals

Ellipse tangents and normals

Hyperbola lines and the changed sign

Continue this lesson.

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