Challenge · Simultaneous Equations (Linear & Quadratic)

No calculator21 marks

These problems combine ideas – you will need to form your own equation (usually a quadratic) and solve it. There is a worked example to show you how to start; then it is over to you. Stuck? Hints are at the foot of the sheet. No calculator.

Worked example: how to start
A line has equation $y=x+3$. It crosses the curve $y=x^2+1$ at two points. Work out the coordinates of both points.
The two $y$-values are equal, so $x^2+1=x+3$.
Rearrange to $x^2-x-2=0$, which factorises as $(x-2)(x+1)=0$, so $x=2$ or $x=-1$.
Substitute back into $y=x+3$: the points are $(2,5)$ and $(-1,2)$.
1The line $y=x+k$ is a tangent to the curve $y=x^2$. Work out the value of $k$.[4 marks]
Answer:
2A circle has equation $x^2+y^2=20$. The line $y=2x$ meets the circle at two points. Work out the coordinates of both points.[3 marks]
Answer:
3A line has gradient $3$ and passes through the point $(0,-5)$. It intersects the curve $y=x^2-3$ at two points, $A$ and $B$. Work out the coordinates of $A$ and $B$, and the coordinates of the midpoint of $AB$.[5 marks]
Answer:
4A straight line passes through the points $(0,1)$ and $(2,5)$. Work out the coordinates of the points where this line meets the curve $y=x^2+1$.[4 marks]
Answer:
5The line $y=x+c$ is a tangent to the circle $x^2+y^2=8$. Work out the two possible values of $c$.[5 marks]
Answer:
★Extension: Show that the line $y=x-3$ never meets the curve $y=x^2+x+1$, whatever point you try.
Stuck? Hints (don't peek unless you need to)1. Set the two expressions equal to get a quadratic in $x$. A tangent means that quadratic has exactly one (repeated) solution, so its discriminant is $0$.2. Substitute $y=2x$ into the circle equation to get an equation in $x$ only.3. First write the equation of the line, then substitute it into the curve.4. Find the equation of the line through the two given points first, then substitute.5. Substitute the line into the circle to get a quadratic in $x$. A tangent gives a repeated root, so set the discriminant equal to $0$.

Solutions & mark scheme · Simultaneous Equations (Linear & Quadratic)

Total: 21 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1The line $y=x+k$ is a tangent to the curve $y=x^2$. Work out the value of $k$.[4]
Model solution
Equate: $x^2=x+k$, i.e. $x^2-x-k=0$.
A tangent gives a repeated root, so the discriminant is $0$: $(-1)^2-4(1)(-k)=0$.
$1+4k=0$, so $k=-\dfrac{1}{4}$.
Answer: k = $\dfrac{-1}{4}$
Marks
✔1Forms $x^2-x-k=0$
✔1Uses discriminant $=0$
✔1$1+4k=0$
✔1$k=-\frac14$
2A circle has equation $x^2+y^2=20$. The line $y=2x$ meets the circle at two points. Work out the coordinates of both points.[3]
Model solution
Substitute: $x^2+(2x)^2=20$, so $5x^2=20$.
$x^2=4$, so $x=2$ or $x=-2$.
Then $y=2x$ gives the points $(2,4)$ and $(-2,-4)$.
Answer: (2, 4) and (-2, -4)
Marks
✔1Substitutes to $5x^2=20$
✔1$x=\pm 2$
✔1Both points $(2,4)$ and $(-2,-4)$
3A line has gradient $3$ and passes through the point $(0,-5)$. It intersects the curve $y=x^2-3$ at two points, $A$ and $B$. Work out the coordinates of $A$ and $B$, and the coordinates of the midpoint of $AB$.[5]
Model solution
The line is $y=3x-5$.
Equate: $x^2-3=3x-5$, so $x^2-3x+2=0$, i.e. $(x-1)(x-2)=0$.
$x=1$ gives $y=-2$; $x=2$ gives $y=1$. So $A(1,-2)$ and $B(2,1)$.
Midpoint $=\left(\dfrac{1+2}{2},\dfrac{-2+1}{2}\right)=(1.5,-0.5)$.
Answer: A(1, -2), B(2, 1), midpoint (1.5, -0.5)
Marks
✔1Line $y=3x-5$
✔1Forms $x^2-3x+2=0$
✔1Solves $x=1,2$
✔1Both points $A(1,-2)$, $B(2,1)$
✔1Midpoint $(1.5,-0.5)$
4A straight line passes through the points $(0,1)$ and $(2,5)$. Work out the coordinates of the points where this line meets the curve $y=x^2+1$.[4]
Model solution
Gradient $=\dfrac{5-1}{2-0}=2$, and it cuts the $y$-axis at $1$, so the line is $y=2x+1$.
Equate: $x^2+1=2x+1$, so $x^2-2x=0$, i.e. $x(x-2)=0$.
$x=0$ gives $y=1$; $x=2$ gives $y=5$. The points are $(0,1)$ and $(2,5)$.
Answer: (0, 1) and (2, 5)
Marks
✔1Gradient $=2$
✔1Line $y=2x+1$
✔1Forms $x^2-2x=0$
✔1Both points $(0,1)$ and $(2,5)$
5The line $y=x+c$ is a tangent to the circle $x^2+y^2=8$. Work out the two possible values of $c$.[5]
Model solution
Substitute: $x^2+(x+c)^2=8$, so $2x^2+2cx+(c^2-8)=0$.
Tangent means the discriminant is $0$: $(2c)^2-4(2)(c^2-8)=0$.
$4c^2-8c^2+64=0$, so $-4c^2+64=0$, giving $c^2=16$.
$c=4$ or $c=-4$.
Answer: c = 4 or c = -4
Marks
✔1Substitutes to $2x^2+2cx+c^2-8=0$
✔1Uses discriminant $=0$
✔1Simplifies to $c^2=16$
✔1$c=4$
✔1$c=-4$
★Extension
If they met, then $x^2+x+1=x-3$.
This rearranges to $x^2+4=0$, i.e. $x^2=-4$.
No real number squares to give $-4$ (the discriminant is $0^2-4(1)(4)=-16<0$), so there is no real solution – the line and curve never meet.
mathedup.co.uk · sheet 98ZG