Areas of Conics

Most of us know area formulas for basic geometrical objects, such as circles, squares, rectangles, and triangles.  (The area of a circle is A = πr², the area of a square is A = s², the area of a rectangle is A = lw, and the area of a triangle is A = ½bh.)  Others of us may even remember learning area formulas of other geometrical objects in high school geometry, such as parallelograms, trapezoids, and equilateral triangles.  (The area of a parallelogram is A = bh, the area of a trapezoid is ½(b₁ + b₂)h, and the area of an equilateral triangle is ¼√3s².)

Other than the circle, the area of conics are neglected in most geometry textbooks, even though the Cartesian equation for each conic are included in most algebra textbooks.  The area of a parabola is A = ²/₃bh (where b and h are the base and height of the rectangle that contains it).

The area of an ellipse is A = πab (where a and b are the major and minor axes of the ellipse). 

The area of a hyperbola is A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|) (where a and b are the axes of the hyperbola and h is the height).
 

The area formulas for the parabola and ellipse are relatively simple, the equation for the hyperbola is not.

To prove the area of a parabola, we can use integration.  The Cartesian equation for a parabola with its vertex at the origin is y = ax², which means another point on the parabola would be (b, ab²). 

The area of a rectangle with opposite vertices on the origin and (b, ab²) is A = b∙ab² = ab³.  The area of the parabola inside this rectangle is A = ∫₀^b ab² – ax² dx = ab²x – ¹/₃ax³] ₀^b = ab²b – ¹/₃ab³ = ab³ – ¹/₃ab³ = ²/₃ab³.  (The left half of the parabola would be the same by symmetry.)  Therefore, the area of the parabola is ²/₃ of the area of the rectangle that contains it, or A = ²/₃bh.

To prove the area of an ellipse, we can compare its Cartesian equation to the Cartesian equation of a circle.  The Cartesian equation for an ellipse centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 + ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(a² – x²), and the Cartesian equation for a circle centered at the origin where a is a radius is x² + y² = a², which rearranged is y = √(a² – x²), so the difference in heights is a factor of ^b/_a. 

    

So if the area of a circle is A = πr², or in this case A = πa², the area of an ellipse must be A = ^b/_aπa² = πab.  (Technically speaking, the area of an ellipse is 4∫₀^{a b}/_a√(a² – x²)dx = ^b/_a 4∫₀^a √(a² – x²)dx = ^b/_aπa² = πab.)

To prove the area of the hyperbola, we can use integration once again.  The Cartesian equation for a horizontal hyperbola centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 – ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(x² – a²). 

The area is then A = 2∫_a^{h+a b}/_a√(x² – a²)dx.  Using a right triangle with legs a and √(x² – a²) and hypotenuse x, tan q = ^{√(x^2 – a^2)}/_a or √(x² – a²) = a tan q, and sec q = ^x/_a or x = a sec q, which means dx = a sec q tan q dq. 

Using substitution, A = 2∫_sec-1(a)^{sec-1(h+a) b}/_a a tan q a sec q tan q dq = 2ab∫₀^{sec-1((h+a)/a)} tan² q sec q dq = 2ab∫₀^{sec-1((h+a)/a)} (sec² q – 1)sec q dq = 2ab∫₀^{sec-1((h+a)/a)} sec³ q – sec q dq.  Integrating results in 2ab[½ sec q tan q + ½ ln|sec q + tan q| – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}  = ab[sec q tan q – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}.  Since sec(sec^-1(x)) = x and tan(sec^-1(x)) = √(x² – 1), A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|).

Although the area of a hyperbola is not an easy formula, both area formulas for a parabola and an ellipse are relatively simple.  The area of an ellipse can also be used to show the special case of a circle where a = b = r (A = πab = π∙r∙r = πr²), and the area formulas of conics nicely correspond with the Cartesian equations of conics learned in algebra.  It makes one wonder why the area formulas for the parabola and ellipse are typically not included in a high school geometry curriculum or textbooks.