TC6, TC7 - Tangents, normals and chords | OCR A-Level Further Mathematics B (MEI)

Find tangent gradients and equations from cartesian, parametric and polar curves. Use CAS derivatives without losing their conditions, handle horizontal and vertical directions, and connect a chord between distinct points with a limiting tangent.

Differentiate the equation at the correct point

An ellipse has two points with the same horizontal coordinate. Can their normals have opposite gradients even though the curve's equation is unchanged?

Investigate the curve defined by $x^2+4y^2=4$ . Solving for $y$ gives two branches, but implicit differentiation lets you find a gradient without choosing and differentiating a square root.

You already know

A line through $(x_0,y_0)$ with finite gradient $m$ has equation $y-y_0=m(x-x_0)$ . A normal is perpendicular to the tangent. Coordinate conversion and a curve's domain must still be respected when locating the point.

When differentiating with respect to $x$ , $y$ varies with $x$ : the derivative of $4y^2$ is $8y\,\mathrm dy/\mathrm dx$ . A CAS implicit-derivative command should give the same chain-rule equation.

Cartesian tangent and normal

At a regular point of $y=f(x)$ , the tangent gradient is $m=f'(x_0)$ . For $m\ne0$ , the normal gradient is

m_N=-\frac1m

A horizontal tangent has a vertical normal; a vertical tangent has a horizontal normal. Use $x=x_0$ or $y=y_0$ directly in those cases.

Not in the formula booklet: learn it.

For an implicit curve, obtain the gradient from its differentiated equation, then check any division you make.

Worked example 1: the upper point

Question. Find the tangent and normal to $x^2+4y^2=4$ at $P=(\sqrt3,1/2)$ .

Plan. Differentiate implicitly, evaluate the tangent gradient, and use the perpendicular gradient for the normal.

\begin{aligned} 2x+8y\frac{\mathrm dy}{\mathrm dx}&=0&&\text{differentiate with respect to }x\\ \frac{\mathrm dy}{\mathrm dx}&=-\frac{x}{4y},\quad y\ne0 &&\text{solve the derivative equation}\\ m_P&=-\frac{\sqrt3}{2}&&\text{substitute }P\\ m_N&=\frac2{\sqrt3}&&\text{negative reciprocal} \end{aligned}

Thus

y-\frac12=-\frac{\sqrt3}{2}(x-\sqrt3) \quad\text{and}\quad y-\frac12=\frac2{\sqrt3}(x-\sqrt3)

Why this step? The point selects the local gradient; a single implicit equation can contain branches with different gradients at the same $x$ .

Answer. The first displayed line is the tangent and the second the normal.

Check. $3+4(1/4)=4$ , so $P$ is on the curve, and $m_Pm_N=-1$ .

Change only the sign of $P$ 's vertical coordinate.

Worked example 2: the reflected lower point

Question. Find the tangent and normal at $Q=(\sqrt3,-1/2)$ on the same curve.

Plan. Keep the implicit derivative and evaluate it at the reflected point.

\begin{aligned} m_Q&=-\frac{\sqrt3}{4(-1/2)}=\frac{\sqrt3}{2} &&\text{new point, same derivative}\\ m_N&=-\frac2{\sqrt3}&&\text{perpendicular gradient}\\ y+\frac12&=\frac{\sqrt3}{2}(x-\sqrt3) &&\text{tangent through }Q\\ y+\frac12&=-\frac2{\sqrt3}(x-\sqrt3) &&\text{normal through }Q \end{aligned}

Why this step? Reflecting in the horizontal axis reverses the sign of every finite gradient, including the normal's.

Answer. The tangent and normal are the last two displayed lines.

Check. Reflect the preceding example by replacing $y$ with $-y$ ; the equations agree.

At $(0,\pm1)$ , the tangent is horizontal and the normal is $x=0$ . At $(\pm2,0)$ , the differentiated equation cannot be solved by division by $y$ ; the local tangent is vertical, $x=\pm2$ , with normal $y=0$ .

Watch out

Wrong: substitute into a reciprocal-gradient formula when the tangent is horizontal or vertical. Right: identify the actual direction and write its coordinate-line equation.

Reflect the point in both axes for a fresh check.

Divide the two parameter derivatives

For $x=x(t)$ , $y=y(t)$ , a small parameter change moves both coordinates. Their rates must be combined; $\mathrm dy/\mathrm dt$ alone is not the tangent gradient in the coordinate plane.

Parametric gradient

At a regular parameter value with $\mathrm dx/\mathrm dt\ne0$ ,

\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}

If $\mathrm dx/\mathrm dt=0$ and $\mathrm dy/\mathrm dt\ne0$ , the tangent is vertical. If the numerator alone is zero, the tangent is horizontal. If both vanish, investigate nearby behaviour before deciding.

Not in the formula booklet: learn it.

Use CAS to differentiate each coordinate with respect to the tracing parameter. For the curve

x=t^2+1,\qquad y=t^3-3t,\qquad t\in\mathbb R,

the output is:

Quantity	CAS derivative
$x'(t)$	$2t$
$y'(t)$	$3t^2-3$
$y'(t)/x'(t)$	$(3t^2-3)/(2t)$ , for $t\ne0$

Worked example 3: a positive parameter

Question. Find the tangent and normal at $t=2$ . Determine whether the normal meets this parametric curve at another real parameter.

Plan. Locate the point, calculate the derivative quotient, then substitute the parameterisation into the normal.

\begin{aligned} (x,y)&=(5,2)&&\text{evaluate the coordinates}\\ m&=\frac{3(2)^2-3}{2(2)}=\frac94 &&\text{quotient of parameter derivatives}\\ y-2&=\frac94(x-5)&&\text{tangent}\\ y-2&=-\frac49(x-5)&&\text{normal}\\ 9(t^3-3t-2)+4(t^2-4)&=0&&\text{intersections with the normal}\\ (t-2)(9t^2+22t+17)&=0&&\text{CAS factorisation} \end{aligned}

Why this step? An equation for the whole normal line can meet the curve elsewhere. Re-substitution finds actual parameters instead of guessing from a plot.

Answer. The tangent and normal are displayed above. The quadratic factor has discriminant $22^2-4(9)(17)=-128<0$ , so $t=2$ is the only real intersection parameter.

Check. Both lines pass through $(5,2)$ , and their gradient product is $-1$ .

Change only $t=2$ to $t=-2$ . The point and gradients reflect in the horizontal axis.

Worked example 4: a negative parameter

Question. Find the tangent and normal at $t=-2$ , and check the other real normal intersections.

Plan. Re-evaluate the same coordinate and derivative formulas before solving the normal-intersection equation.

\begin{aligned} (x,y)&=(5,-2)&&\text{new parameter point}\\ m&=\frac{3(-2)^2-3}{2(-2)}=-\frac94 &&\text{negative denominator}\\ y+2&=-\frac94(x-5)&&\text{tangent}\\ y+2&=\frac49(x-5)&&\text{normal}\\ 9(t^3-3t+2)-4(t^2-4)&=0&&\text{normal intersections}\\ (t+2)(9t^2-22t+17)&=0&&\text{CAS factorisation} \end{aligned}

Why this step? Squaring the parameter in $x$ does not make the whole parameterisation even; the ordinate changes sign.

Answer. The displayed tangent and normal; $t=-2$ is the sole real normal intersection, since the remaining quadratic again has discriminant $-128$ .

Check. Reflection of the preceding pair of lines gives exactly these equations.

At $t=0$ , the non-zero vertical rate gives a vertical tangent at $(1,0)$ . At $t=\pm1$ , the horizontal rates are non-zero but the vertical rates vanish, giving horizontal tangents.

Watch out

Wrong: use $3t^2-3$ as the coordinate-plane gradient. Right: divide by $2t$ , and handle its zero separately.

Move only the parameter to an exceptional horizontal-tangent value.

Use the polar derivative with its coordinate directions

For a polar curve, differentiate $x=r\cos\theta$ and $y=r\sin\theta$ . This connects the polar formula to the parametric method you have just used.

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TC6, TC7 - Tangents, normals and chords

Differentiate the equation at the correct point

Divide the two parameter derivatives

Use the polar derivative with its coordinate directions

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