Porisma

Worked solution

An ellipse with equal radii

Write down the equation of the ellipse shown in the diagram.

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An ellipse-shaped closed curve has centre (h,k). A leftwards horizontal radius and a downwards vertical radius are both labelled r. Use these equal lengths to determine its equation.
An ellipse-shaped closed curve has centre (h,k). A leftwards horizontal radius and a downwards vertical radius are both labelled r. Use these equal lengths to determine its equation.
An ellipse-shaped closed curve has centre (h,k). A leftwards horizontal radius and a downwards vertical radius are both labelled r. Use these equal lengths to determine its equation.

CJC · Lecture · Example 3

Read the two radii

The general centred form is (x−h)2a2+(y−k)2b2=1\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1. Both labelled semi-axes in the diagram are rr, so a=b=ra=b=r. The curve is a circle, a special case of an ellipse.

How did this part go?

Write the equation

(x−h)2r2+(y−k)2r2=1\dfrac{(x-h)^2}{r^2}+\dfrac{(y-k)^2}{r^2}=1. Multiplying by r2r^2 gives (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2, the circle with centre (h,k)(h,k) and radius rr. The source’s concluding remark is that a circle is an ellipse with equal axes.

How did this part go?

Common mistakes

Use r² in each denominator. Equal semi-axis lengths describe a circle; the orientation of the drawn radius segments does not change its equation.

Solution notes

Student and Teacher copies, PDF page 8, printed page 120. Both show the same completed derivation and equal-axes remark. No marks are printed.

Only symbolic centre and radius labels are given; the numerical placement and size in the reconstructed diagram are illustrative.

Read the hint