NSWMaths Extension 1Syllabus dot point

When a line meets a curve, what does it mean for the line to be a tangent rather than a secant, and how can the sum and product of the roots of one equation locate the point of contact, the midpoint of a chord, and a common tangent without any calculus?

Apply the factor theorem, multiplicity and the sum and product of roots to the geometry of curves: a line is tangent to a curve exactly when the equation formed by solving them simultaneously has a double root; the x-coordinate of the midpoint of a chord is the average of the roots; and the number and nature of the intersections of two curves are read from the roots of the difference of the two polynomials

Geometry of curves through polynomial roots for HSC Maths Extension 1. A line is tangent exactly when solving line and curve gives a double root; a chord midpoint is the average of the roots; intersections are read from the difference of the polynomials. Tangents, chords and common tangents without calculus.

Generated by Claude Opus 4.820 min answerUpdated 2026-06-21

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Quick answer

When a line meets a curve $y = P(x)$ , solving them simultaneously gives one equation $P(x) - (mx + k) = 0$ whose roots are the x-coordinates of the meeting points. Two distinct roots mean the line cuts the curve (a secant); a double (repeated) root means the line touches it (a tangent), and the repeated root is the point of contact. For a quadratic intersection equation the tangent condition is the discriminant $\Delta = b^2 - 4ac = 0$ , with contact at $x = -\tfrac{b}{2a}$ ; for a cubic, set the roots with a repeat and use the sum and product of roots. The midpoint of a chord has x-coordinate equal to the average of the roots, $x_M = \tfrac{1}{2}(\alpha + \beta)$ , read straight off the coefficients without solving for the roots. From a point outside a parabola there are two tangents: write the line with unknown gradient $m$ , set the discriminant of the intersection equation to zero, and solve the resulting quadratic in $m$ . When a point lies on a cubic, its x-coordinate is automatically a root, so the sum of the other two roots is fixed and the chord midpoint lies on a fixed vertical line for every gradient; the tangent through the point arises when those two roots merge. For two general curves, factor the difference $P(x) - Q(x)$ : a double factor is a point of tangency (a touch, where the curves share a common tangent) and a simple factor is a crossing. Every result comes from roots, the discriminant and multiplicity, with no calculus.

What this dot point is asking

This is the payoff page for the whole polynomials module. Every tool built so far, the factor theorem, multiplicity, and the sum and product of roots, turns out to answer questions about the geometry of curves, and to answer them more cleanly than the brute-force algebra most students reach for. The single idea that unlocks the page is this: when you solve a line and a curve simultaneously you get one polynomial equation, and the geometry of how they meet is written in the roots of that equation. Two simple roots mean the line cuts the curve at two points (a secant). A double root means the line just touches it (a tangent). The midpoint of a chord is the average of the roots, which the sum of roots hands you for free.

NESA groups this under ME-F2 as the geometric application of polynomial techniques. You should be able to find where a line is tangent to a curve using the double-root or discriminant condition, find the two tangents from an external point, handle a line through a fixed point on a cubic that meets it twice more, locate the midpoint of the resulting chord, and find a tangent common to two curves. Crucially, this is the Year 11 treatment: every result here comes from roots, the discriminant and multiplicity, never from differentiating to get a gradient. The calculus route to tangents waits for Year 12; here the polynomial route is the whole lesson, and it is often faster. (Cambridge gates this material as "rather demanding, could be Enrichment", which is exactly why a careful, fully worked treatment beats the textbook.)

The answer

Solving two curves gives one equation whose roots are the meeting points

Take any two curves $y = P(x)$ and $y = Q(x)$ . They meet exactly where $P(x) = Q(x)$ , that is where the single equation

P(x) - Q(x) = 0

holds. This is the master move of the whole page: two curves collapse into one polynomial, and the x-coordinates of every intersection are the roots of that polynomial. When one of the curves is a line $y = mx + k$ , the equation is $P(x) - (mx + k) = 0$ , and its roots are the x-coordinates of the points where the line meets the curve. Everything that follows is just reading those roots carefully.

The degree of $P(x) - Q(x)$ caps how many times the curves can meet, exactly as on the consequences of the factor theorem page: a line ( $\deg 1$ ) meets a parabola ( $\deg 2$ ) at most twice and a cubic ( $\deg 3$ ) at most three times, because that is the most roots the difference can have.

Tangency means a double root

Here is the central insight. A line is a secant when it cuts a curve at two separate points, and a tangent when it touches the curve at a single point without crossing. In root language:

two distinct roots of $P(x) - (mx + k) = 0$ are two separate crossing points: the line is a secant;
a double (repeated) root is a single point where the line touches but does not cross: the line is a tangent, and the repeated root is the x-coordinate of the point of contact.

The reason is the multiplicity rule from earlier in the module. Near a simple root the difference $P(x) - (mx + k)$ changes sign, so the curve passes from one side of the line to the other (a crossing). Near a double root the difference has a repeated factor $(x - \alpha)^2$ , which never goes negative, so the curve returns to the side it came from without crossing: it touches. A tangent is the line for which two intersection points have merged into one repeated contact.

The midpoint of a chord is the average of the roots

The second pillar is just as useful. Suppose a line cuts a curve at two points $A$ and $B$ whose x-coordinates are $\alpha$ and $\beta$ , the two roots of the intersection equation. The midpoint $M$ of the chord $AB$ has x-coordinate

x_M = \frac{\alpha + \beta}{2},

the average of the roots, and its y-coordinate is found by putting $x_M$ into the line (since $M$ , like $A$ and $B$ , lies on the line). The point of this is that the sum $\alpha + \beta$ comes straight from the coefficients, with no need to solve for $\alpha$ and $\beta$ individually. For a quadratic intersection equation $ax^2 + bx + c = 0$ ,

x_M = \frac{\alpha + \beta}{2} = \frac{1}{2}\left(-\frac{b}{a}\right) = -\frac{b}{2a}.

This is why so many exam questions can be answered "without finding $A$ and $B$ ": the midpoint never needed the individual roots, only their sum.

Two tangents from an external point

From a point outside a parabola there are exactly two tangent lines, and the double-root condition finds both at once. Write the general line through the external point with unknown gradient $m$ , form the quadratic intersection equation, and set its discriminant to zero. The discriminant condition is itself a quadratic in $m$ , and its two solutions are the gradients of the two tangents. (A point inside the parabola gives a negative discriminant in $m$ , hence no real tangents; a point on the parabola gives a repeated $m$ , the single tangent there.) Each gradient then gives a point of contact from the repeated root $x = \dfrac{m}{2a}$ of the intersection equation.

A line through a fixed point on a cubic

A favourite Extension 1 set-up: a point $P$ lies on a cubic, and a line of variable gradient $m$ through $P$ meets the cubic at two further points $A$ and $B$ . Because $P$ is on both the line and the curve, its x-coordinate is automatically a root of the cubic intersection equation, for every $m$ . That leaves a clean structure: the three roots are (x-coordinate of $P$ ), $\alpha$ , $\beta$ , and the sum of all three is fixed by the coefficient of $x^2$ . So $\alpha + \beta$ , and therefore the midpoint of $AB$ , is pinned down regardless of $m$ . The midpoint sweeps a single vertical line as $m$ varies.

Tangency arises as the limiting case. As the line tilts, $A$ and $B$ slide along the curve; when they merge ( $\alpha = \beta$ ) the line becomes tangent at that doubled point, distinct from $P$ . Setting $\alpha = \beta$ and using the known value of $\alpha + \beta$ fixes the contact point, and the product of roots then gives the special gradient $m$ . This single picture contains both the chord-midpoint result and the tangent, and it is the clearest demonstration of why the sum and product of roots are the right tools.

Where two curves touch and where they cross

The same difference $P(x) - Q(x)$ handles two general curves, not just a line and a curve. Factor the difference: each double factor is a point where the curves are tangent to each other (they touch), and each simple factor is a point where they cross. This is the curve-to-curve version of the tangency rule, and one difference polynomial can encode several meetings of different kinds at once. The example below pairs a cubic and a parabola whose difference is $(x - 2)^2(x + 1)$ : tangent at the double root $x = 2$ , crossing at the simple root $x = -1$ .

When two curves touch at a point, they are tangent to each other there, and a single line is tangent to both at that point: a common tangent. Finding the value of a parameter that makes two curves touch is therefore the standard way to set up a common-tangent problem without calculus, as in the worked examples below.

Exam technique

These questions reward a tidy, root-focused layout. To bank the marks:

Always start by solving the two curves into one equation $P(x) - Q(x) = 0$ , and write "the x-coordinates of the points of intersection satisfy ...". That sentence frames the whole solution and earns the first mark.
For a tangent, say the magic words. Write "the line is a tangent, so the roots are equal" (or "so $\Delta = 0$ "). For a quadratic use $\Delta = b^2 - 4ac = 0$ ; for a cubic, set the roots as $\theta, \theta, \dots$ and use the sum and product. Examiners look for the explicit double-root justification, not just the answer.
Use the sum of roots for a midpoint; do not solve for the roots. State $x_M = \tfrac{1}{2}(\alpha + \beta)$ and read $\alpha + \beta$ off the coefficients. Substitute into the line for $y_M$ .
When a given point lies on the curve, exploit the known root. Its x-coordinate is automatically a root, so the remaining roots have a sum and product fixed by the coefficients, which is what locates the chord midpoint and the tangent.
Find the contact point from the repeated root. For a quadratic, contact is at $x = -\tfrac{b}{2a}$ ; for a forced double root $\alpha$ , contact is at $x = \alpha$ . Then read $y$ off the line (or either curve) and confirm with the other.
Sketch the situation. A quick sketch of the curve, the line and the contact or chord keeps the geometry straight and catches sign slips. NESA explicitly expects sketches on these questions.
No calculus needed. Reaching for the derivative is a Year 12 method; here the roots, the discriminant and multiplicity do everything, and usually faster.

How exam questions ask about geometry using polynomials

The wordings map directly onto the root tool to reach for.

"Show that the x-coordinates of the points of intersection satisfy [equation]." Set $P(x) = Q(x)$ and rearrange to one polynomial equal to zero. This is almost always the first part.
"Show that the line is a tangent / find the value of [the constant] for which the line is a tangent." Impose a double root: $\Delta = 0$ for a quadratic intersection equation, or equal roots via sum and product for a cubic.
"Find the point of contact." It is the repeated root: $x = -\tfrac{b}{2a}$ for a quadratic, or the doubled root $\alpha$ for a cubic; then find $y$ from the line.
"Find the coordinates of the midpoint $M$ of the chord." Take $x_M = \tfrac{1}{2}(\alpha + \beta)$ from the sum of roots, then $y_M$ from the line. Do not solve for $\alpha$ and $\beta$ .
"Two tangents are drawn from [an external point]; find them." Line through the point with gradient $m$ , set the discriminant of the intersection quadratic to zero, get a quadratic in $m$ with two solutions.
"A line through [a point on the curve] meets the curve again at $A$ and $B$ ; show $M$ lies on [a fixed line]." Use the known root and the sum of the other two roots, which is fixed by the coefficients.
"Show the two curves touch / find [a parameter] so the curves are tangent / find the common tangent." Factor the difference $P(x) - Q(x)$ ; a double factor is a point of tangency, and the shared tangent line is found there.

Worked example

Find where a line of given gradient is tangent to a parabola

Find the value of $b$ for which the line $y = 2x + b$ is a tangent to the parabola $y = x^2 - 4x + 7$ , and find the point of contact $T$ .

Solve the line and the parabola simultaneously. Points of intersection satisfy $x^2 - 4x + 7 = 2x + b$ . Bringing all terms to one side,

x^2 - 6x + (7 - b) = 0.

Impose the tangent condition as a double root. The line is a tangent exactly when this quadratic has a repeated root, that is when its discriminant is zero. With $a = 1$ , $\,\text{coefficient of } x = -6$ , constant $7 - b$ ,

\Delta = (-6)^2 - 4(1)(7 - b) = 36 - 28 + 4b = 8 + 4b.

Setting $\Delta = 0$ gives $8 + 4b = 0$ , so $b = -2$ .

Find the point of contact from the repeated root. With $b = -2$ the equation is $x^2 - 6x + 9 = 0$ , that is $(x - 3)^2 = 0$ , so the double root is $x = 3$ . The contact point lies on the line: $y = 2(3) - 2 = 4$ . Hence $T = (3, 4)$ .

Check. On the parabola, $3^2 - 4(3) + 7 = 9 - 12 + 7 = 4$ , agreeing with $T$ . The perfect square $(x - 3)^2$ is the verified double root, confirming tangency. So $y = 2x - 2$ is tangent at $T(3, 4)$ .

Find the two tangents to a parabola from an external point

Find the equations of the two tangents from the point $P(2, 1)$ to the parabola $y = x^2$ , and the points of contact.

Write a general line through $P$ . A non-vertical line through $P(2, 1)$ has the form $y - 1 = m(x - 2)$ , that is $y = mx + (1 - 2m)$ .

Form the intersection equation. Meeting the parabola where $x^2 = mx + (1 - 2m)$ gives

x^2 - mx + (2m - 1) = 0.

Impose tangency. The line is a tangent when this quadratic has equal roots, so

\Delta = (-m)^2 - 4(1)(2m - 1) = m^2 - 8m + 4 = 0.

This is a quadratic in $m$ ; its two solutions are the gradients of the two tangents. By the quadratic formula,

m = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(4)}}{2} = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3}.

Write the two tangents. Substituting each $m$ into $y = mx + (1 - 2m)$ ,

y = (4 + 2\sqrt{3})x + (1 - 2(4 + 2\sqrt{3})) = (4 + 2\sqrt{3})x - 7 - 4\sqrt{3},

y = (4 - 2\sqrt{3})x - 7 + 4\sqrt{3}.

Find the points of contact. For each tangent the contact x-coordinate is the repeated root $x = \dfrac{m}{2}$ , so $x = 2 \pm \sqrt{3}$ , and $y = x^2$ gives the contact heights:

x = 2 + \sqrt{3} \implies y = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3}, \qquad x = 2 - \sqrt{3} \implies y = 7 - 4\sqrt{3}.

State and check. The two tangents touch $y = x^2$ at $\left(2 + \sqrt{3},\ 7 + 4\sqrt{3}\right)$ and $\left(2 - \sqrt{3},\ 7 - 4\sqrt{3}\right)$ . As a check, each tangent passes through $P(2, 1)$ : at $x = 2$ , $y = m(2) + 1 - 2m = 1$ for either gradient, as required. Two real tangents confirm that $P$ lies outside the parabola.

A line through a point on a cubic: chord midpoint and the tangent

The point $P(2, 0)$ lies on the cubic $y = x^3 - 4x$ . A line of gradient $m$ through $P$ meets the cubic at two further points $A(\alpha, y_A)$ and $B(\beta, y_B)$ .

Set up the intersection equation. The line through $P(2, 0)$ is $y = m(x - 2)$ . It meets the cubic where $x^3 - 4x = m(x - 2)$ , that is

x^3 - (4 + m)x + 2m = 0.

Putting $x = 2$ gives $8 - 2(4 + m) + 2m = 8 - 8 - 2m + 2m = 0$ , so $x = 2$ is a root for every $m$ , as it must be because $P$ lies on both the line and the curve.

Locate the chord midpoint from the sum of roots. The three roots are $2$ , $\alpha$ and $\beta$ . There is no $x^2$ term, so the sum of all three roots is $0$ :

2 + \alpha + \beta = 0 \implies \alpha + \beta = -2.

The midpoint $M$ of $AB$ has x-coordinate $x_M = \dfrac{\alpha + \beta}{2} = -1$ , for every $m$ . So as the line tilts, the midpoint of the chord always sits on the vertical line $x = -1$ .

Force tangency by merging $A$ and $B$ . The line becomes tangent to the cubic at a point other than $P$ when $A$ and $B$ coincide, that is $\alpha = \beta$ . Combined with $\alpha + \beta = -2$ this gives $\alpha = \beta = -1$ .

Find the gradient from the product of roots. For $x^3 - (4 + m)x + 2m$ the product of the roots is $-\dfrac{2m}{1} = -2m$ . With roots $2$ , $-1$ , $-1$ ,

2 \cdot (-1) \cdot (-1) = 2 = -2m \implies m = -1.

State and check the double root. The tangent through $P$ is $y = -(x - 2) = -x + 2$ , and the point of contact is at $x = -1$ : on the cubic $y = (-1)^3 - 4(-1) = 3$ , so the contact is $(-1, 3)$ . With $m = -1$ the intersection equation is $x^3 - 3x - 2 = 0$ , which factors as $(x + 1)^2(x - 2) = 0$ ; the repeated factor $(x + 1)^2$ is the verified double root, confirming tangency at $(-1, 3)$ .

A common tangent line tangent to two curves at the same point

Show that the cubic $y = x^3$ and the parabola $y = 3x^2 - 4$ touch at one point and cross at another, and find the common tangent line at the point where they touch.

Form the difference of the two polynomials. The curves meet where $x^3 = 3x^2 - 4$ , that is where $F(x) = 0$ for

F(x) = x^3 - (3x^2 - 4) = x^3 - 3x^2 + 4.

Factor the difference. The constant term is $4$ , so test divisors $\pm 1, \pm 2, \pm 4$ . $F(-1) = -1 - 3 + 4 = 0$ , so $x + 1$ is a factor; $F(2) = 8 - 12 + 4 = 0$ , so $x - 2$ is a factor. Accounting for the cubic (the constant term $4 = (-2)^2 \cdot 1$ shows $2$ is the repeat),

F(x) = (x - 2)^2 (x + 1).

Read off touch and cross. The double root $x = 2$ is a point where the curves are tangent to each other (they touch), and the simple root $x = -1$ is a point where they cross. So the curves touch at $x = 2$ and cross at $x = -1$ , as claimed.

Find the common tangent at the touch point. At $x = 2$ the curves meet at $y = 2^3 = 8$ , so the point of contact is $(2, 8)$ . The shared tangent there is the line tangent to the cubic at $(2, 8)$ ; to find its gradient $m$ without calculus, require the line $y = m(x - 2) + 8$ to be tangent to the cubic, that is $x^3 - [m(x - 2) + 8] = 0$ to have a double root at $x = 2$ . Now

x^3 - mx + 2m - 8 = (x - 2)\big(x^2 + 2x + (4 - \tfrac{m}{2} \cdot 0)\big)

is awkward to keep general, so instead match a double factor directly. Writing $x^3 - mx + (2m - 8) = (x - 2)^2(x - r)$ and expanding the right side gives $x^3 - (4 + r)x^2 + (4 + 4r)x - 4r$ . The coefficient of $x^2$ on the left is $0$ , so $4 + r = 0$ , giving $r = -4$ . Then the coefficient of $x$ gives $-m = 4 + 4r = 4 - 16 = -12$ , so $m = 12$ , and the constant gives $2m - 8 = -4r = 16$ , that is $m = 12$ , agreeing.

State and check. The common tangent at the touch point is $y = 12(x - 2) + 8 = 12x - 16$ . Check the double root: with $m = 12$ , $x^3 - 12x + 16 = (x - 2)^2(x + 4)$ , confirming the line touches the cubic at $(2, 8)$ . The same line is tangent to the parabola there: $3x^2 - 4 = 12x - 16$ gives $3x^2 - 12x + 12 = 3(x - 2)^2 = 0$ , a double root at $x = 2$ as well. So $y = 12x - 16$ is the common tangent at $(2, 8)$ .

Find a parameter so a parabola touches a quartic

Find the value of $k$ for which the parabola $y = x^2 + k$ touches the quartic $y = x^4$ at two points, and find the points of contact.

Form the difference of the two polynomials. The curves meet where $x^4 = x^2 + k$ , that is where $F(x) = 0$ for

F(x) = x^4 - x^2 - k.

Encode "touches at two points" as two double roots. If the parabola touches the quartic at two points, the difference $F(x)$ has a double root at each, so it is a perfect square of a quadratic,

F(x) = (x^2 - t)^2 = x^4 - 2t x^2 + t^2

for some constant $t$ . (Only even powers appear because $F$ is even, which fits the symmetry of two contact points at $\pm$ the same x-value.)

Match coefficients. Comparing $x^4 - x^2 - k$ with $x^4 - 2t x^2 + t^2$ :

-2t = -1 \implies t = \tfrac{1}{2}, \qquad t^2 = -k \implies k = -t^2 = -\tfrac{1}{4}.

Find the points of contact. The double roots are where $x^2 = t = \tfrac{1}{2}$ , so $x = \pm\dfrac{1}{\sqrt{2}}$ . The contact height is $y = x^4 = \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ .

State and check. With $k = -\dfrac{1}{4}$ the parabola $y = x^2 - \tfrac{1}{4}$ touches $y = x^4$ at $\left(\dfrac{1}{\sqrt{2}},\ \dfrac{1}{4}\right)$ and $\left(-\dfrac{1}{\sqrt{2}},\ \dfrac{1}{4}\right)$ . Check the difference: $x^4 - x^2 + \tfrac{1}{4} = \left(x^2 - \tfrac{1}{2}\right)^2$ , a perfect square, so both contacts are genuine double roots and the curves touch (rather than cross) at each.

Common traps

Using a single root instead of a double root for a tangent: A tangent does not mean "the line meets the curve once". A vertical-ish line or a special direction can also meet a parabola once. The precise condition is a repeated root of the intersection equation, that is $\Delta = 0$ for a quadratic, not merely "one solution".
Forgetting that an external point gives two tangents: Setting $\Delta = 0$ produces a quadratic in $m$ . Solve it fully: there are two gradients, hence two tangents. Stopping at one tangent loses half the marks.
Solving for the roots when only their sum is needed: For a chord midpoint you need $\alpha + \beta$ , which the coefficients give directly. Grinding out the (often irrational) roots wastes time and risks slips; quote the sum-of-roots relation instead.
Dropping the known root on a "point on the curve" problem: When $P$ lies on the curve, its x-coordinate is one root of the intersection equation for every value of the gradient. Use it: it is what makes the sum of the remaining roots, and hence the midpoint, constant.
Reaching for calculus: Differentiating to find the tangent gradient is a Year 12 technique and is not the intended Year 11 method here. The discriminant, the sum and product of roots, and multiplicity answer every part, and usually more quickly. (If you do use calculus to check, make sure the written solution still gives the polynomial-root reasoning the marks are awarded for.)
Confusing a curve-curve touch with a crossing in the difference: In $P(x) - Q(x)$ , a squared factor is a tangency (touch) and a simple factor is a crossing. Misreading the multiplicity reverses the geometry. Factor carefully and match the constant term to pin down which root repeats.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFind the value of

c

for which the line

y = 6x + c

is a tangent to the parabola

y = x^2 + 2x + 5

, and find the point of contact.

Show worked solution →

Set the line equal to the curve. A tangent meets the parabola where $x^2 + 2x + 5 = 6x + c$ . Bringing everything to one side,

x^2 - 4x + (5 - c) = 0.

Apply the double-root (tangent) condition. The line is a tangent exactly when this quadratic has a repeated root, that is when the discriminant is zero. With $a = 1$ , $b = -4$ , $C = 5 - c$ ,

\Delta = (-4)^2 - 4(1)(5 - c) = 16 - 20 + 4c = 4c - 4.

Setting $\Delta = 0$ gives $4c - 4 = 0$ , so $c = 1$ .

Find the point of contact. With $c = 1$ the equation is $x^2 - 4x + 4 = 0$ , that is $(x - 2)^2 = 0$ , so the double root is $x = 2$ . The contact point lies on the line: $y = 6(2) + 1 = 13$ .

State and check. The line is $y = 6x + 1$ , tangent at $(2, 13)$ . Check on the parabola: $2^2 + 2(2) + 5 = 4 + 4 + 5 = 13$ . The double root confirms tangency.

foundation3 marksShow that the line

y = -2x + 4

is a tangent to the parabola

y = x^2 - 6x + 8

, and find the point of contact

T

Show worked solution →

Form the equation of intersection. The line and parabola meet where $x^2 - 6x + 8 = -2x + 4$ . Collecting terms,

x^2 - 4x + 4 = 0.

Show the root is repeated. The left side factors as a perfect square,

x^2 - 4x + 4 = (x - 2)^2,

so the only solution is the double root $x = 2$ . (Equivalently $\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$ .) A repeated root means the line meets the curve at exactly one point and does not cross, which is precisely what it means to be a tangent.

Find the point of contact. At $x = 2$ the line gives $y = -2(2) + 4 = 0$ , so $T = (2, 0)$ .

Check on the parabola. $2^2 - 6(2) + 8 = 4 - 12 + 8 = 0$ , agreeing. So $y = -2x + 4$ is tangent at $T(2, 0)$ .

core4 marksThe line

y = x + 4

meets the parabola

y = x^2 - 2x + 1

at two points

A

and

B

. Without finding

A

and

B

, find the coordinates of the midpoint

M

of the chord

AB

Show worked solution →

Form the equation whose roots are the x-coordinates. The points of intersection occur where $x^2 - 2x + 1 = x + 4$ , that is

x^2 - 3x - 3 = 0.

Its two roots $\alpha$ and $\beta$ are the x-coordinates of $A$ and $B$ . (The discriminant is $(-3)^2 - 4(1)(-3) = 9 + 12 = 21 > 0$ , so there are indeed two distinct points: a genuine chord.)

Use the sum of the roots for the midpoint x-coordinate. For $x^2 - 3x - 3$ the sum of the roots is $\alpha + \beta = -\dfrac{-3}{1} = 3$ . The midpoint's x-coordinate is the average of the two x-coordinates,

x_M = \frac{\alpha + \beta}{2} = \frac{3}{2}.

Find the midpoint y-coordinate. Both $A$ and $B$ lie on the line $y = x + 4$ , and the midpoint of a chord of a line lies on that line, so

y_M = x_M + 4 = \frac{3}{2} + 4 = \frac{11}{2}.

State the answer. The midpoint is $M = \left(\dfrac{3}{2},\ \dfrac{11}{2}\right)$ , found from the sum of the roots alone, without solving the quadratic.

core5 marksThe cubic

y = x^3 - 7x + 6

passes through

P(1, 0)

. A line of gradient

m

through

P

meets the curve at two further points

A

and

B

. (a) Show that the x-coordinates of

A

and

B

have sum

-1

, so the midpoint of

AB

lies on the line

x = -\tfrac{1}{2}

for every

m

. (b) Find the value of

m

for which the line is tangent to the curve at a point other than

P

, and find that point of contact.

Show worked solution →

Form the intersection equation. The line through $P(1, 0)$ with gradient $m$ is $y = m(x - 1)$ . It meets the cubic where $x^3 - 7x + 6 = m(x - 1)$ , that is

x^3 - (7 + m)x + (6 + m) = 0.

Substituting $x = 1$ gives $1 - (7 + m) + (6 + m) = 0$ , so $x = 1$ is a root for every $m$ , as it must be since $P$ lies on both curves.

(a) Use the sum of the roots. Let the three roots be $1$ , $\alpha$ and $\beta$ . There is no $x^2$ term, so the sum of all three roots is $0$ :

1 + \alpha + \beta = 0 \implies \alpha + \beta = -1.

The midpoint of $AB$ has x-coordinate $\dfrac{\alpha + \beta}{2} = -\dfrac{1}{2}$ , independent of $m$ . So $M$ always lies on the vertical line $x = -\dfrac{1}{2}$ .

(b) Impose tangency by merging the two further points. The line is tangent at a point other than $P$ when $A$ and $B$ coincide, that is $\alpha = \beta$ . With $\alpha + \beta = -1$ this forces $\alpha = \beta = -\dfrac{1}{2}$ .

Use the product of the roots to find $m$ . For $x^3 - (7 + m)x + (6 + m)$ the product of the roots is $-\dfrac{6 + m}{1} = -(6 + m)$ . With roots $1$ , $-\tfrac{1}{2}$ , $-\tfrac{1}{2}$ ,

1 \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{1}{2}\right) = \frac{1}{4} = -(6 + m) \implies m = -\frac{25}{4}.

Find the contact point and confirm the double root. At $x = -\tfrac{1}{2}$ the cubic gives $y = \left(-\tfrac{1}{2}\right)^3 - 7\left(-\tfrac{1}{2}\right) + 6 = -\tfrac{1}{8} + \tfrac{7}{2} + 6 = \tfrac{75}{8}$ . With $m = -\tfrac{25}{4}$ the intersection equation is $x^3 - \tfrac{3}{4}x - \tfrac{1}{4} = 0$ , which factors as $\left(x + \tfrac{1}{2}\right)^2 (x - 1) = 0$ : the repeated factor confirms the tangency. So $m = -\dfrac{25}{4}$ and the point of contact is $\left(-\dfrac{1}{2},\ \dfrac{75}{8}\right)$ .

exam6 marksFrom the external point

P(0, -8)

, two tangents are drawn to the parabola

y = x^2 - 4

. Find the equations of the two tangents and their points of contact.

Show worked solution →

Set up a general line through $P$ . A non-vertical line through $P(0, -8)$ has y-intercept $-8$ , so it is $y = mx - 8$ for some gradient $m$ .

Form the intersection equation. This line meets the parabola where $x^2 - 4 = mx - 8$ , that is

x^2 - mx + 4 = 0.

Apply the tangent condition. The line is a tangent when this quadratic has a double root, so its discriminant is zero:

\Delta = (-m)^2 - 4(1)(4) = m^2 - 16 = 0 \implies m = 4 \ \text{ or } \ m = -4.

Two values of $m$ give the two tangents, as expected from a point outside the parabola.

Find each point of contact. For a double root the contact x-coordinate is $x = \dfrac{m}{2}$ (the repeated root of $x^2 - mx + 4$ ).

$m = 4$ : the equation is $x^2 - 4x + 4 = (x - 2)^2 = 0$ , contact at $x = 2$ , $y = 2^2 - 4 = 0$ , so $(2, 0)$ .
$m = -4$ : the equation is $x^2 + 4x + 4 = (x + 2)^2 = 0$ , contact at $x = -2$ , $y = (-2)^2 - 4 = 0$ , so $(-2, 0)$ .

State and check. The tangents are $y = 4x - 8$ and $y = -4x - 8$ , touching the parabola at $(2, 0)$ and $(-2, 0)$ respectively. Both pass through $P$ : $4(0) - 8 = -8$ and $-4(0) - 8 = -8$ . Each contact is a confirmed double root.

exam5 marksShow that the cubic

y = x^3 + x

and the parabola

y = -2x^2 + 8x - 4

touch at one point and cross at another. Find both points and state the nature of each.

Show worked solution →

Form the difference of the two polynomials. The curves meet where $x^3 + x = -2x^2 + 8x - 4$ , that is where $F(x) = 0$ for

F(x) = (x^3 + x) - (-2x^2 + 8x - 4) = x^3 + 2x^2 - 7x + 4.

Factor the difference. The constant term is $4$ , so test divisors $\pm 1, \pm 2, \pm 4$ . $F(1) = 1 + 2 - 7 + 4 = 0$ , so $x - 1$ is a factor; $F(-4) = -64 + 32 + 28 + 4 = 0$ , so $x + 4$ is a factor. Dividing accounts for the cubic as

F(x) = (x - 1)^2 (x + 4),

with $x = 1$ found as a double root (matching the constant term $(-1)^2(4) = 4$ ).

Read off the nature of each meeting: A double root means the curves are tangent (touch without crossing); a simple root means they cross. So the curves are tangent at $x = 1$ and cross at $x = -4$ .
Find the coordinates: At $x = 1$ : $y = 1^3 + 1 = 2$ , so the tangency is at $(1, 2)$ . At $x = -4$ : $y = (-4)^3 + (-4) = -64 - 4 = -68$ , so the crossing is at $(-4, -68)$ .
State and check: The curves touch at $(1, 2)$ (double root) and cross at $(-4, -68)$ (simple root). Check the parabola at $x = 1$ : $-2 + 8 - 4 = 2$ , agreeing; and at $x = -4$ : $-32 - 32 - 4 = -68$ , agreeing.

What this dot point is asking

The answer

Solving two curves gives one equation whose roots are the meeting points

Tangency means a double root

The midpoint of a chord is the average of the roots

Two tangents from an external point

A line through a fixed point on a cubic

Where two curves touch and where they cross

How exam questions ask about geometry using polynomials

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