9709 Pure 1 · 1.3 Coordinate Geometry · Session 3 of 4

Where Lines Meet Curves

Does a line cut a curve, touch it, or miss it completely? Today one move answers that for circles, parabolas and reciprocal curves, and it is the same discriminant you used in 1.1.

Do Now, and why this matters7 min

Know these curve shapes5 min

One move for every line and curve, and the Meeting Lab13 min

Finding k for a tangent, where it touches, and a 5-mark Cambridge question16 min

Spot the mistake, then on your own12 min

Recap and exit ticket5 min

Do Now · 5 min · no notes, no calculator

Where does the line \(y = x + 2\) meet the circle \(x^2 + y^2 = 20\)?

Substitute, solve, and give both points. Then find \(b^2 - 4ac\) for the quadratic you solved.

Write all of it, then tap to check.

replace \(y\) with \(x + 2\): \(x^2 + (x + 2)^2 = 20\)

expand, \((x + 2)^2 = x^2 + 4x + 4\): \(2x^2 + 4x + 4 = 20\), so \(2x^2 + 4x - 16 = 0\)

divide by 2 and factor: \(x^2 + 2x - 8 = 0\), \((x + 4)(x - 2) = 0\), \(x = 2\) or \(x = -4\)

\(y = x + 2\) for each: \(x = 2\) gives \(y = 4\); \(x = -4\) gives \(y = -2\)

discriminant of \(x^2 + 2x - 8\): \(2^2 - 4(1)(-8) = 4 + 32 = 36\)

\((2, 4)\) and \((-4, -2)\); \(b^2 - 4ac = 36\). Two roots, two meeting points. Hold on to that.

Why this matters · 2 min

A line and a curve can cross twice, touch once, or never meet.

Each meeting point is a pair \((x, y)\) that fits both equations at once. So "where do they meet?" is the same question as "solve these together."

What each skill answers

Substitute and solve: exactly where do the paths cross?

The discriminant: how many times do they meet, without solving?

Finding k for a tangent: what setting makes a path just graze the edge?

Where you will see it

Boats near a reef. A protected zone around a reef off St. Croix is close to a circle. Does a boat's straight course cross into it, graze its edge, or stay clear?

Video games. Does a shot's straight path hit a round target? Game engines solve this same quadratic many times a second, and the discriminant says hit or miss.

Roads. A straight road that just grazes a roundabout meets it at one point: a tangent.

Light. A laser beam hitting a curved lens or mirror. Where it lands is a line meeting a curve.

Your Do Now found two meeting points, and \(b^2 - 4ac\) was positive. By the end you can predict cut, touch or miss before you solve anything, which is what Cambridge asks.

Direct instruction · 1 of 6 · know these curve shapes

Sketch the curve first. Six shapes to know.

\(y = x\)
a line

\(y = x^2\)
a parabola

\(y = x^3\)
a cubic

\(y = \tfrac{1}{x}\)
a reciprocal

\(y = \tfrac{1}{x^2}\)
both branches up

\(y = \sqrt{x}\)
starts at 0

\(y = kx^n\), \(k > 0\) \(k < 0\): flipped in the \(x\)-axis. The size of \(k\) only stretches it. Reciprocals get closer and closer to the axes but never reach them. Those lines are called asymptotes.

Predict: which quadrants is \(y = -\dfrac{1}{2x}\) in? Tap to check.

it is \(\tfrac{1}{x}\) times \(-\tfrac{1}{2}\): flipped in the \(x\)-axis and squashed to half height

2nd and 4th quadrants. For \(x > 0\), \(y\) is negative; for \(x < 0\), \(y\) is positive. You will need this picture in 15 minutes.

Direct instruction · 2 of 6 · one move for every line and curve

Substitute, rearrange to \(\ldots = 0\), then read \(b^2 - 4ac\).

\(x^2 + y^2 = 20\) cuts touches misses

Why it works: each root of the quadratic is the \(x\)-coordinate of a meeting point. So the number of roots is the number of meeting points, and \(b^2 - 4ac\) counts the roots.

\(y = x + 2\), your Do Now: \(x^2 + 2x - 8 = 0\), \(b^2 - 4ac = 36 > 0\)
two roots: the line cuts the circle twice

\(y = 2x + 10\): \(x^2 + (2x + 10)^2 = 20\) gives \(5x^2 + 40x + 80 = 0\), so \(x^2 + 8x + 16 = 0\)
\(8^2 - 4(1)(16) = 0\) one repeated root, \(x = -4\): it touches at \((-4, 2)\)

\(y = 2x + 13\): \(5x^2 + 52x + 149 = 0\)
\(52^2 - 4(5)(149) = 2704 - 2980 = -276 < 0\) no real roots: it misses

\(b^2 - 4ac > 0\): cuts, two points · \(= 0\): touches, a tangent · \(< 0\): misses

Rearrange to \(\ldots = 0\) first. \(a\), \(b\) and \(c\) only mean anything once everything is on one side.

Two more things. You can substitute for \(x\) instead: you get a quadratic in \(y\), and \(b^2 - 4ac\) works the same. And this test needs a quadratic. For a cubic, a tangent can touch the curve at one point and still cut it somewhere else.

Direct instruction · 3 of 6 · the Meeting Lab

Pick a curve, set the line, predict, then show it.

the curve the line \(y = mx + c\) the slip: \(\tfrac{2}{x}\) read as \(2x\)

Direct instruction · 4 of 6 · finding k for a tangent

For which values of \(k\) is \(y = 3x + k\) a tangent to \(x^2 + y^2 = 10\)?

\(x^2 + y^2 = 10\) \(y = 3x + 10\) and \(y = 3x - 10\)

Picture check: every line \(y = 3x + k\) has gradient 3, so they are all parallel. A circle has two parallel tangents, one on each side. Expect two answers.

1 · substitute \(y = 3x + k\): \(x^2 + (3x + k)^2 = 10\)

2 · expand, \((3x + k)^2 = 9x^2 + 6kx + k^2\), and rearrange to 0:
\(10x^2 + 6kx + k^2 - 10 = 0\) so \(a = 10\), \(b = 6k\), \(c = k^2 - 10\)

3 · tangent means touches once, so \(b^2 - 4ac = 0\):
\((6k)^2 - 4(10)(k^2 - 10) = 0\)

4 · simplify: \(36k^2 - 40k^2 + 400 = 0\), so \(-4k^2 + 400 = 0\), \(k^2 = 100\)

\(k = 10\) or \(k = -10\)

Do both survive? The question puts no condition on \(k\), and the picture has two tangents, so keep both. Next: where does each line touch the circle?

Direct instruction · 5 of 6 · where does the tangent touch?

Found \(k\)? Put it back into the quadratic to find where the line touches.

The move. You never guess \(x\). Put each value of \(k\) back into the quadratic from step 2. Because \(b^2 - 4ac = 0\), it is always a perfect square, so it gives exactly one \(x\): the touching point. Then get \(y\) from the line.

Quicker: when \(b^2 - 4ac = 0\), the one root is \(x = -\dfrac{b}{2a}\)

Why it works. Start from the quadratic formula:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(b^2 - 4ac = 0\): \(x = \dfrac{-b \pm 0}{2a} = -\dfrac{b}{2a}\)

Adding or taking away 0 changes nothing, so the two roots are the same number.

When you can use it: only when \(b^2 - 4ac = 0\) (a tangent). Use \(a\) and \(b\) from the quadratic rearranged to \(\ldots = 0\), after \(k\) is put in. If \(b^2 - 4ac > 0\), \(-\tfrac{b}{2a}\) is halfway between the two roots, not a meeting point. It gives \(x\) only; \(y\) still comes from the line.

\(k = 10\): put it into \(10x^2 + 6kx + k^2 - 10 = 0\)
\(10x^2 + 60x + 90 = 0\) ÷ 10: \(x^2 + 6x + 9 = 0\)
\((x + 3)^2 = 0\), \(x = -3\) or quicker: \(x = -\dfrac{60}{2(10)} = -3\)
from the line: \(y = 3(-3) + 10 = 1\)

\(k = -10\): put it in the same way
\(10x^2 - 60x + 90 = 0\) ÷ 10: \(x^2 - 6x + 9 = 0\)
\((x - 3)^2 = 0\), \(x = 3\) or quicker: \(x = -\dfrac{-60}{2(10)} = 3\)
from the line: \(y = 3(3) - 10 = -1\)

\(y = 3x + 10\) touches at \((-3, 1)\); \(y = 3x - 10\) touches at \((3, -1)\)

Check: \((-3)^2 + 1^2 = 10\) and \(3^2 + (-1)^2 = 10\). Both points are on the circle.

Direct instruction · 6 of 6 · June 2023, Paper 11, Q5 · 5 marks

The line \(y = kx - k\), where \(k\) is a positive constant, is a tangent to the curve \(y = -\dfrac{1}{2x}\).

Find, in either order, the value of \(k\) and the coordinates of the point where the tangent meets the curve. [5]

\(y = -\tfrac{1}{2x}\) \(y = 2x - 2\)

You met this question in 1.1. Now you can see it: the curve is in the 2nd and 4th quadrants, and the line touches one branch. "In either order" means you may find \(k\) or the point first.

What the examiners saw: many students found this hard. \(-\tfrac{1}{2x}\) often became \(-\tfrac{1}{2}x\), which is not a quadratic at all. Some kept \(k = 0\). When a quadratic gives two answers, check both make sense.

*M1 · where they meet, both give the same \(y\), so set them equal: \(kx - k = -\dfrac{1}{2x}\)

clear the fraction: multiply both sides by \(2x\) (fine, the curve has no point at \(x = 0\)), then add 1:
\(2kx^2 - 2kx = -1\) \(\Rightarrow\) \(2kx^2 - 2kx + 1 = 0\) a quadratic: \(a = 2k\), \(b = -2k\), \(c = 1\)

DM1 · tangent, so \(b^2 - 4ac = 0\):
\((-2k)^2 - 4(2k)(1) = 0\), \(4k^2 - 8k = 0\), \(4k(k - 2) = 0\)

A1 · \(k = 0\) or \(k = 2\). Reject 0, since \(k\) is positive: \(k = 2\) only

B1 · put \(k = 2\) back: \(4x^2 - 4x + 1 = 0\), \((2x - 1)^2 = 0\), \(x = \tfrac{1}{2}\)
or quicker, since \(b^2 - 4ac = 0\): \(x = -\tfrac{b}{2a} = \tfrac{4}{8} = \tfrac{1}{2}\)

B1 · from the line: \(y = 2\left(\tfrac{1}{2}\right) - 2 = -1\)

\(k = 2\), point \(\left(\tfrac{1}{2}, -1\right)\)

Spot the mistake · 3 min

Three starts on Q5. What is wrong with each?

\(y = kx - k\), \(k > 0\), is a tangent to \(y = -\dfrac{1}{2x}\). A and B are real exam mistakes.

A. "\(kx - k = -\tfrac{1}{2}x\), so \(\left(k + \tfrac{1}{2}\right)x = k\)"

\(-\tfrac{1}{2x}\) has \(x\) on the bottom; \(-\tfrac{1}{2}x\) has it on top. Different curves.

Multiply both sides by \(2x\). Then you get a quadratic: \(2kx^2 - 2kx + 1 = 0\).

B. "\(4k(k - 2) = 0\), so \(k = 0\) or \(k = 2\). Two tangents."

\(k\) must be positive. Also, \(k = 0\) gives \(y = 0\) (the \(x\)-axis), which the curve never touches.

why 0 appeared: with \(k = 0\), \(2kx^2 - 2kx + 1 = 0\) becomes \(1 = 0\). No \(x^2\) term, so \(b^2 - 4ac\) never applied.

Only \(k = 2\): one tangent.

C. "Tangent, so \(b^2 - 4ac > 0\)"

\(> 0\) means two roots: the line cuts the curve twice.

A tangent touches once: \(b^2 - 4ac = 0\).

Practice · on your own · 9 min, finish at home

Sketch first. Show the quadratic and \(b^2 - 4ac\). Tap to check.

1. The line \(y = mx - 3\) is a tangent to \(y = x^2 - 5x + 6\). Find \(m\).

\(x^2 - 5x + 6 = mx - 3\) \(\Rightarrow\) \(x^2 - (5 + m)x + 9 = 0\); tangent: \((5 + m)^2 - 36 = 0\), \(5 + m = \pm 6\)

\(m = 1\) or \(m = -11\)

where each touches, put \(m\) back:
\(m = 1\): \(x^2 - 6x + 9 = 0\), \((x - 3)^2 = 0\)
\(x = 3\), \(y = 3 - 3 = 0\)
\(m = -11\): \(5 + m = -6\), so \(x^2 + 6x + 9 = 0\)
\((x + 3)^2 = 0\), \(x = -3\), \(y = -11(-3) - 3 = 30\)
or quicker, \(x = -\tfrac{b}{2a} = \tfrac{5 + m}{2}\): \(3\) and \(-3\)

two answers, two different lines: \(y = x - 3\) touches at \((3, 0)\); \(y = -11x - 3\) touches at \((-3, 30)\). Each is a tangent.

2. Find the set of values of \(k\) for which the line \(y = kx + 1\) does not meet the curve \(y = x^2 + 4\).

\(x^2 + 4 = kx + 1\) \(\Rightarrow\) \(x^2 - kx + 3 = 0\)

does not meet, so \(b^2 - 4ac < 0\): \(k^2 - 12 < 0\)

critical values \(\pm\sqrt{12} = \pm 2\sqrt{3}\), and \(< 0\) is the between region (1.1):

\(-2\sqrt{3} < k < 2\sqrt{3}\)

Practice · on your own · question 3

A "show that" question: prove it for every \(k\), not just one.

\(y = \tfrac{2}{x}\) \(y = x + k\) for \(k = -3\), \(0\), \(3\)

Slide the line up or down as far as you like: it still crosses both branches. The algebra proves this for every \(k\).

3. Show that the line \(y = x + k\) meets the curve \(y = \dfrac{2}{x}\) at two points, whatever the value of \(k\).

"show that ... whatever \(k\)" means: do not find \(k\). Prove \(b^2 - 4ac > 0\) for every possible \(k\).

set equal, multiply by \(x\), rearrange: \(x^2 + kx - 2 = 0\), \(a = 1\), \(b = k\), \(c = -2\)

\(b^2 - 4ac = k^2 - 4(1)(-2) = k^2 + 8\) (minus times minus is plus)

\(k^2\) is a square, never negative (\(k = -3\) gives 9, \(k = 0\) gives 0), so \(k^2 + 8\) is at least 8.

\(k^2 + 8 > 0\) for every \(k\), so the line always cuts the curve at two points.

Recap and homework · 2 min

Three things to carry out of here.

Line meets curve: substitute, rearrange to \(\ldots = 0\), read \(b^2 - 4ac\). Positive: cuts twice. Zero: touches, a tangent. Negative: misses.

Finding \(k\) for a tangent: set \(b^2 - 4ac = 0\) and solve for \(k\). Check every value against the question and the picture. Then put each \(k\) back to find where it touches, or use \(x = -\tfrac{b}{2a}\).

Sketch the curve first. Fractions like \(-\tfrac{1}{2x}\): multiply through by the bottom. It is not \(-\tfrac{1}{2}x\).

Homework, as posted in Schoology. Finish practice questions 1 to 3 if you did not get there. Stuck? Write down where you got stuck and bring it; that still counts as done.

Next lesson: chords, the angle in a semicircle, and the rest of the June 2023 circle question, then a short 1.3 review. Review day is Tuesday, November 3, and Test 1 is Thursday, November 5. Intersection Lab

Exit ticket · 5 min · Schoology, Paper 1 → 1.3 Coordinate Geometry

Four questions. Work on paper first, then enter your answers in Schoology.

1. Substitute \(y = x + 1\) into \(x^2 + y^2 = 25\) and rearrange to \(\ldots = 0\). Which equation do you get?

2. The line \(y = 3x + c\) is a tangent to the curve \(y = x^2 + 5x + 2\). Find \(c\).

3. Does the line \(y = x - 1\) cut, touch or miss the curve \(y = x^2 - 3x + 3\)? If it meets the curve, where?

4. The line \(y = kx + 2\) is a tangent to the curve \(y = -\dfrac{3}{x}\). Find \(k\) and the point where the line touches the curve. Show the quadratic and \(b^2 - 4ac\).

Submit before you leave. An attempt left open scores nothing.

If time · extra practice

Two more. Watch which values of \(k\) survive.

The line \(y = kx - 5\) is a tangent to the curve \(y = x^2 - x - 1\). Find the possible values of \(k\) and the point where each tangent touches.

\(x^2 - x - 1 = kx - 5\) \(\Rightarrow\) \(x^2 - (1 + k)x + 4 = 0\)

\(b^2 - 4ac = 0\): \((1 + k)^2 - 16 = 0\), \(1 + k = \pm 4\), \(k = 3\) or \(k = -5\)

\(k = 3\): \(x^2 - 4x + 4 = 0\), \(x = 2\), \(y = 3(2) - 5 = 1\) · \(k = -5\): \(x^2 + 4x + 4 = 0\), \(x = -2\), \(y = -5(-2) - 5 = 5\)

\(k = 3\), touching at \((2, 1)\); \(k = -5\), touching at \((-2, 5)\). Nothing in the question rules either out, so both stay.

Find the set of values of \(k\) for which the line \(y = kx - 1\) meets the curve \(y = x^2 + 3x + 3\) at two distinct points.

\(x^2 + 3x + 3 = kx - 1\) \(\Rightarrow\) \(x^2 + (3 - k)x + 4 = 0\)

two distinct points, so \(b^2 - 4ac > 0\): \((3 - k)^2 - 16 > 0\)

critical values: \(3 - k = \pm 4\), so \(k = -1\) or \(k = 7\). \(> 0\) is the outside region:

\(k < -1\) or \(k > 7\). Compare practice Q2: "does not meet" gave a between region.