A line meeting a parabola and a circle

Geometry, grade 10 (5 units) Lesson 5 of 26Learning

A line meeting a parabola and a circle

A line and a parabola

The meeting points of a line and a parabola are the solutions of the system of their equations. Substituting the line’s $\displaystyle y$ in the parabola’s equation leaves a quadratic equation in $\displaystyle x$ .

Example: the parabola $\displaystyle {y}={x^{2}}$ and the line $\displaystyle {y}={x+2}$ .

$\displaystyle {x^{2}}={x+2}$ , so $\displaystyle {x^{2}-x-2}={0}$ and $\displaystyle {(x-2)(x+1)}={0}$ ;
$\displaystyle {x}={2}$ : $\displaystyle {y}={4}$ ; $\displaystyle {x}={-1}$ : $\displaystyle {y}={1}$ ;
the meeting points: $\displaystyle (2,4)$ and $\displaystyle (-1,1)$ .

\displaystyle y=x^{2}

\displaystyle y=x+2

\displaystyle A=(2,4)

\displaystyle B=(-1,1)

Exercise 1

The parabola: $\displaystyle {y}={x^{2}-2x-3}$ ; the line: $\displaystyle {y}={x+1}$ .

Find the meeting points by solving the system.

Mark the meeting points on the board.

\displaystyle y=x^{2}-2x-3

\displaystyle y=x+1

Two meeting points, one or none

The number of meeting points is the number of solutions of the quadratic equation, and its discriminant $\displaystyle {\Delta}={b^{2}-4ac}$ tells it:

$\displaystyle \Delta>0$ – two meeting points: the line cuts the parabola;
$\displaystyle {\Delta}={0}$ – one meeting point: a line that is not parallel to the parabola’s axis is then a tangent – it touches the parabola;
$\displaystyle \Delta<0$ – no meeting point.

Example: the line $\displaystyle {y}={2x+k}$ and the parabola $\displaystyle {y}={x^{2}}$ .

$\displaystyle {x^{2}}={2x+k}$ , so $\displaystyle {x^{2}-2x-k}={0}$ ;
$\displaystyle {\Delta}={(-2)^{2}-4\cdot1\cdot(-k)}={4+4k}$ ;
$\displaystyle {4+4k}={0}$ at $\displaystyle {k}={-1}$ : the line $\displaystyle {y}={2x-1}$ is a tangent to the parabola, at $\displaystyle {x}={1}$ , the point $\displaystyle (1,1)$ .

\displaystyle y=x^{2}

\displaystyle y=2x-1

\displaystyle y=2x+2

\displaystyle y=2x-3

\displaystyle T=(1,1)

Exercise 2

The parabola: $\displaystyle {y}={x^{2}+3}$ ; the line: $\displaystyle {y}={kx-1}$ .

Substitute and arrange:

\displaystyle {x^{2}-kx+4}={0}

. For which

\displaystyle k

is the discriminant

\displaystyle 0

For which values of $\displaystyle k$ does the line cut the parabola at two points?
For which does it not meet the parabola at all?

A line and a circle

A line and a circle meet at two points, at one – the line is then a tangent to the circle – or at none. Their meeting points are again the solutions of a system.

Example: the circle $\displaystyle {(x-1)^{2}+(y+1)^{2}}={25}$ and the line $\displaystyle {y}={x-1}$ .

$\displaystyle {(x-1)^{2}+(x-1+1)^{2}}={25}$ , so $\displaystyle {x^{2}-2x+1+x^{2}}={25}$ ;
$\displaystyle {2x^{2}-2x-24}={0}$ , or $\displaystyle {x^{2}-x-12}={0}$ , and $\displaystyle {(x-4)(x+3)}={0}$ ;
$\displaystyle {x}={4}$ : $\displaystyle {y}={3}$ ; $\displaystyle {x}={-3}$ : $\displaystyle {y}={-4}$ ;
the meeting points: $\displaystyle (4,3)$ and $\displaystyle (-3,-4)$ . The segment between them is a chord of the circle.

\displaystyle (x-1)^{2}+(y+1)^{2}=25

\displaystyle y=x-1

\displaystyle A=(4,3)

\displaystyle B=(-3,-4)

Exercise 3

The circle: $\displaystyle {x^{2}+y^{2}}={25}$ ; the line: $\displaystyle {y}={x+1}$ .

Find the meeting points.

Find the length of the chord the circle cuts off the line.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

Does the line $\displaystyle {y}={-x+8}$ meet the circle $\displaystyle {x^{2}+y^{2}}={16}$ ?
Answer by solving the system.

Exercise 5

The line $\displaystyle {x}={k}$ is parallel to the $\displaystyle y$ -axis.
For which $\displaystyle k$ is it a tangent to the circle $\displaystyle {(x-2)^{2}+(y-1)^{2}}={9}$ ?

\displaystyle (x-2)^{2}+(y-1)^{2}=9

Formulas: geometry, grade 10 (5 units)

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Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$