Where a Line Meets a Curve | CAIE 9709 Pure Mathematics 1 Notes

Chapters

Pure Mathematics 1

Mathematics9709

Substitute, then solve

make one variable the subjectsubstitute into the curvesolve the quadraticsubstitute back for both points

Worked example

Find where $y=2x-1$ meets $y=x^2-4$

Set the two expressions for $y$ equal.
$x^2-4=2x-1$
$x^2-2x-3=0$
$(x+1)(x-3)=0$
Put each $x$ back into the line to get its $y$ .
$x=-1 \Rightarrow y=-3$
$x=3 \Rightarrow y=5$

Answer

(-1,\,-3) \quad\text{and}\quad (3,\,5)

Common mistake

Do not stop at the

x

-values. A point needs both coordinates; substitute back into the linear equation.

Worked example

Find the points of intersection of $2xy+5y^2=24$ and $2x+y+4=0$ (9709/12/M/J/25 Q2)

Make $2x$ the subject because the curve contains $2xy$ .
$2x=-y-4$
$(-y-4)y+5y^2=24$
$y^2-y-6=0$
$(y-3)(y+2)=0$
Here the first line was simplified and divided by $4$ . Then find each $x$ from the line.

Answer

\left(-\tfrac{7}{2},\,3\right) \quad\text{and}\quad (-1,\,-2)

Your turn9709/11/M/J/25 Q6(a)

The equation of a curve is $2x^2-kxy+2=0$ and the equation of a line is $y=px+3$ . Given that $k=2$ and $p=11$ , find the coordinates of the points of intersection.

Show worked answer

2x^2-2x(11x+3)+2=0

Expand, collect the terms and divide by $-2$ .

10x^2+3x-1=0

(5x-1)(2x+1)=0

Answer

\left(\tfrac{1}{5},\,\tfrac{26}{5}\right) \quad\text{and}\quad \left(-\tfrac{1}{2},\,-\tfrac{5}{2}\right)

Cut, touch or miss

Swipe left or right to see the whole diagram →

Three lines of the same gradient against y = x². Solving the two equations together gives a quadratic, and its discriminant says which picture you are in.

\Delta>0

→ two points

\Delta=0

→ touches once (tangent)

\Delta<0

→ no meeting

Substitute, then solve

make one variable the subjectsubstitute into the curvesolve the quadraticsubstitute back for both points

Worked example

Find where $y=2x-1$ meets $y=x^2-4$

Set the two expressions for $y$ equal.
$x^2-4=2x-1$
$x^2-2x-3=0$
$(x+1)(x-3)=0$
Put each $x$ back into the line to get its $y$ .
$x=-1 \Rightarrow y=-3$
$x=3 \Rightarrow y=5$

Answer

(-1,\,-3) \quad\text{and}\quad (3,\,5)

Common mistake

Do not stop at the

x

-values. A point needs both coordinates; substitute back into the linear equation.

Worked example

Find the points of intersection of $2xy+5y^2=24$ and $2x+y+4=0$ (9709/12/M/J/25 Q2)

Make $2x$ the subject because the curve contains $2xy$ .
$2x=-y-4$
$(-y-4)y+5y^2=24$
$y^2-y-6=0$
$(y-3)(y+2)=0$
Here the first line was simplified and divided by $4$ . Then find each $x$ from the line.

Answer

\left(-\tfrac{7}{2},\,3\right) \quad\text{and}\quad (-1,\,-2)

Your turn9709/11/M/J/25 Q6(a)

The equation of a curve is $2x^2-kxy+2=0$ and the equation of a line is $y=px+3$ . Given that $k=2$ and $p=11$ , find the coordinates of the points of intersection.

Show worked answer

2x^2-2x(11x+3)+2=0

Expand, collect the terms and divide by $-2$ .

10x^2+3x-1=0

(5x-1)(2x+1)=0

Answer

\left(\tfrac{1}{5},\,\tfrac{26}{5}\right) \quad\text{and}\quad \left(-\tfrac{1}{2},\,-\tfrac{5}{2}\right)

Cut, touch or miss

Swipe left or right to see the whole diagram →

Three lines of the same gradient against y = x². Solving the two equations together gives a quadratic, and its discriminant says which picture you are in.

\Delta>0

→ two points

\Delta=0

→ touches once (tangent)

\Delta<0

→ no meeting