Solve Inequalities Involving Ellipses
When we solve an inequality, we typically expect a range of solutions. This is the same for when we solve inequalities involving ellipses, with the exception that the solutions satisfying the inequality is a region on the $xy$xy-plane.
Exploration
Consider the graph of the following ellipse $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=1$(x−3)252+(y+2)232=1.
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| Graph of the ellipse $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=1$(x−3)252+(y+2)232=1 |
We can think of the graph as the set of all points that satisfy the equation $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=1$(x−3)252+(y+2)232=1.
In light of this, we might be curious in finding the set of all points that satisfies the inequality
$\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}<1$(x−3)252+(y+2)232<1
or with any inequality symbol for that matter.
We know that the set of points satisfying the equation $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=\frac{1}{4}$(x−3)252+(y+2)232=14 is certainly going to satisfy the inequality since the result of the left-hand side is a quarter, which is less than one. Let's rewrite this equation into a more familiar form.
| $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}$(x−3)252+(y+2)232 | $=$= | $\frac{1}{4}$14 | (Equation of the ellipse) |
| $\frac{\left(x-3\right)^2}{5^2\times\frac{1}{4}}+\frac{\left(y+2\right)^2}{3^2\times\frac{1}{4}}$(x−3)252×14+(y+2)232×14 | $=$= | $1$1 | (Dividing both sides by $\frac{1}{4}$14) |
| $\frac{\left(x-3\right)^2}{5^2\times\left(\frac{1}{2}\right)^2}+\frac{\left(y+2\right)^2}{3^2\times\left(\frac{1}{2}\right)^2}$(x−3)252×(12)2+(y+2)232×(12)2 | $=$= | $1$1 | (Rewriting $\frac{1}{4}$14 with a power) |
| $\frac{\left(x-3\right)^2}{2.5^2}+\frac{\left(y+2\right)^2}{1.5^2}$(x−3)22.52+(y+2)21.52 | $=$= | $1$1 | (Using the properties of powers) |
The semi-major and semi-minor axes are $2.5$2.5 units and $1.5$1.5 units respectively, but the centre of the ellipse remains the same, $\left(3,-2\right)$(3,−2). Drawing this graph, we can see that it is enclosed by the original ellipse $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=1$(x−3)252+(y+2)232=1.
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| Graph of the ellipse $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=1$(x−3)252+(y+2)232=1 containing the graph of the ellipse $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=\frac{1}{4}$(x−3)252+(y+2)232=14. |
In fact, we can rewrite the equation of the ellipse $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=r$(x−3)252+(y+2)232=r for any value of $r$r that is less than $1$1, such as $r=\frac{1}{2},\frac{1}{3},\frac{4}{5},...$r=12,13,45,..., and draw the resulting set of concentric ellipses.
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| Graph of several ellipses satisfying $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}<1$(x−3)252+(y+2)232<1 |
Ultimately, if we could draw all the ellipses satisfying the equation $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}=r$(x−3)252+(y+2)232=r where $r<1$r<1, then we would obtain the shaded region below.
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| Set of points satisfying $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}<1$(x−3)252+(y+2)232<1 |
This is exactly the set of points satisfying the inequality $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}<1$(x−3)252+(y+2)232<1. We dash the ellipse to indicate that the shaded region excludes the points that are actually on the ellipse itself.
Using the same procedure, the set of points satisfying the inequality $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}>1$(x−3)252+(y+2)232>1 may be represented by the shaded region outside of the ellipse as shown below.
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| Set of points satisfying $\frac{\left(x-3\right)^2}{5^2}+\frac{\left(y+2\right)^2}{3^2}>1$(x−3)252+(y+2)232>1 |
We summarise the regions satisfying inequalities involving ellipses below.
Consider the ellipse $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2+(y−k)2b2=1 where $a
- The set of points satisfying $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}<1$(x−h)2a2+(y−k)2b2<1 lie inside the ellipse.
- The set of points satisfying $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}>1$(x−h)2a2+(y−k)2b2>1 lie outside the ellipse.
If the inequality is not strict ($\le$≤ or $\ge$≥) then the set of points satisfying the inequality also include the points on the ellipse as well.
Note that the above summary also holds if we swap the order of $a$a and $b$b in the equation of the ellipse.
Worked example
Does the point $\left(0,0\right)$(0,0) lie inside, outside or on the ellipse $\frac{\left(x+5\right)^2}{4^2}+\frac{\left(y-8\right)^2}{6^2}=1$(x+5)242+(y−8)262=1?
Think: We know that the ellipse is centred at $\left(-5,8\right)$(−5,8) and has a horizontal semi-minor axis of $4$4 units.
Do: So the ellipse is bound between the lines $x=-9$x=−9 and $x=-1$x=−1, which means the point $\left(0,0\right)$(0,0) lies outside of the ellipse.
Reflect: Alternatively we can substitute the point $\left(0,0\right)$(0,0) into left-hand side of the equation. We expect the result to be greater than $1$1 since the point lies outside the ellipse.
Practice questions
question 1
Which of these points lie inside the ellipse $\frac{x^2}{8^2}+\frac{y^2}{3^2}=1$x282+y232=1? Select all that apply.
$\left(7,0\right)$(7,0)
A$\left(0,0\right)$(0,0)
B$\left(0,7\right)$(0,7)
C$\left(-7,0\right)$(−7,0)
D
question 2
Which of the following points lie inside, outside or on the ellipse $\frac{x^2}{8^2}+\frac{\left(y-5\right)^2}{11^2}=1$x282+(y−5)2112=1?
$\left(0,5\right)$(0,5)
Inside the ellipse
AOutside the ellipse
BOn the ellipse
C$\left(0,0\right)$(0,0)
Inside the ellipse
AOutside the ellipse
BOn the ellipse
C$\left(2,10\right)$(2,10)
Inside the ellipse
AOutside the ellipse
BOn the ellipse
C$\left(2\sqrt{14},0\right)$(2√14,0)
Inside the ellipse
AOutside the ellipse
BOn the ellipse
C
question 3
Which shaded region represents the set of points satisfying the inequality $\frac{x^2}{5^2}+\frac{y^2}{9^2}\le1$x252+y292≤1?
A
B
C
D