All parabolas through three non-collinear points

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1 ALL PARABOLAS THROUGH THREE NON-COLLINEAR POINTS 03 All parabolas through three non-collinear points STANLEY R. HUDDY and MICHAEL A. JONES If no two of three non-collinear points share the same -coordinate, then the parabola = a + a + a 0 through the points is easil found b solving a sstem of linear equations. That is but one of an infinite number of parabolas through the three points. How does one find the other parabolas? In this note, we find all parabolas through an three noncollinear points b reducing the problem to finding the equation of a parabola b using rotations. The parabola = a + a + a 0 has an ais of smmetr parallel to the -ais. Other parabolas have an ais of smmetr that is parallel to some line = m. We focus on the angle θ that the ais of smmetr makes with the -ais, as in Figure, so that tan θ = /m. To find the parabola associated with θ that goes through three non-collinear points, we rotate the three points counterclockwise b θ, find the equation of the parabola, and then rotate the parabola (and the three points) counterclockwise back b θ so that the parabola goes through the original points. As we eplain below, this approach shows that there are three angles for which there are no parabolas through the three points and allows us to find the parabola with the widest opening, or widest parabola, through the points. m θ θ m = m = m m < 0 m > 0 FIGURE : θ ( π /, 0) (Left); θ (0, π /) (Right). A motivating eample To find the parabola through the points (, 3), (, ) and (0, 0) and with ais of smmetr parallel to =, we rotate the points around the origin counterclockwise b θ = π /4 because m = and tan θ = /m. cos θ sin θ The points are rotated b the rotation matri R θ = for sin θ cos θ θ = π /4. The origin (0, 0) remains at the origin after rotation, but the points (, 3) and (, ) become, respectivel,, D5 / C C: C 5 D

2 04 THE MATHEMATICAL GAZETTE 3 = 5 and =. 0 Now, we solve for the equation of the parabola = a + a + a 0 through the rotated points b solving the sstem of equations in the a i when the rotated points are substituted into the equation of the parabola. Because the parabola goes through the point (0, 0), it follows that a 0 = 0. The other two points result in two linear equations and two unknowns: ( ) a + ( ) a = ( ) a + ( ) a = 0. 5 Solving these equations simultaneousl gives a = 0 and a = 5. The resulting parabola = 5 0 goes through the rotated points, as shown in Figure (Left). For the net step, we will refer to this parabola as Pand b the equation f (, ) = 5 0 = 0. (, 5 ) 4 4 (, 3) and (, ) (,0) (0, 0) (0, 0) FIGURE : The parabola = 5 0 through the points ( /, 5 /), (, 0) and (0, 0) (Left); The parabola = 0 through the points (, 3), (, ) and (0, 0) (Right) Finall, the parabola P given b f (, ) = 0 is rotated back b θ, using the rotation matri R θ. This rotates the points back to their initial positions and rotates the parabola as well. To generate the parabola through the original points, P is rotated counterclockwise b θ (i.e. clockwise b θ) which ields the desired parabola P. A point (, ) lies on the parabola P if, and onl if, R θ (, ) lies on P. (This follows because R θ (P) = P, so that P = R θ (P ).) Hence the points R θ (, ) = ( cos θ sin θ, sin θ + cos θ) must satisf f (, ) = 0, or, equivalentl, the parabola P is the set of (, ) for which f ( cos θ sin θ, sin θ + cos θ) = 0. For the eample where θ = π /4, then the parabola P is the set of (, ) for which f (( /) ( /), ( /) +( /) ) = 0, which is equivalent to, D5 / C C: C 5 D

3 ALL PARABOLAS THROUGH THREE NON-COLLINEAR POINTS 05 5 [( /) ( /) ] 0 [( /) ( /) ] ( /) ( /) = 0. This simplifies to = 0. This parabola P has ais of smmetr parallel to = and goes through the original three points; see Figure (Right).. The general problem Suppose we want to find a parabola with an ais of smmetr parallel to = m and through the non-collinear points (, ), (, ) and ( 3, 3 ). We ma assume that ( 3, 3 ) = (0, 0) ; if not, then we can solve the problem for ( 3, 3 ), ( 3, 3 ) and (0, 0) and then translate the resulting parabola b ( 3, 3 ). The advantage of using (0, 0) for the third point is that (0, 0) remains fied upon rotation; this ensures that the constant term a 0 of the parabolic equation = a + a + a 0 is 0. In the introduction, we noted that if an of the three non-collinear points shared an -value, then no parabola goes through the three points. From this observation, it is clear that there are three aes of smmetr for which there is no parabola through the three points. These aes have slopes parallel to the sides of the triangle made from the three points, as rotating the points b the associated angles results in two of the three points sharing -values. For the initial points (, ), (, ) and (0, 0), a rotation b θ i = tan (/m i ) when the ais of smmetr has a slope of m =, m = 0 0, or m 3 = 0 0 results in two of the rotated points having the same -values. The angles for i = to 3 are pictured in Figure 3. θ i (, ) m 3 θ (, ) θ 3 θ m FIGURE 3: The angles θ i (0,0) for which there are no parabolas We now consider the general problem for non-collinear points (, ), (, ) and (0, 0) and a desired ais of smmetr with slope m where m m i for the m i as defined above for i = to 3. Let θ = tan (/m). First, we rotate the points using the rotation matri. Recall that (0, 0) is, D5 / C C: C 5 D R θ

4 06 THE MATHEMATICAL GAZETTE fied under rotation. Rotating (, ) and (, ) gives R θ i i = cos θ sin θ sin θ cos θ i i = cos θ i sin θ i sin θ i + cos θ i The rotated points (cos θ i sin θ i, sin θ i + cos θ i ), for i =,, are substituted into the equation = a + a (because a 0 = 0) to ield two equations and two unknowns: sin θ i + cos θ i = a [cos θ i sin θ i ] +a [cos θ i sin θ i ], for i =,. The two equations can be written as t = a s + a s and t = a s + a s, where t = sin θ + cos θ t = sin θ + cos θ and s = cos θ sin θ s = cos θ sin θ. Solving this sstem of equations for a and a ields a = s t s t s s (s s ) and a = It follows that the parabola through the rotated points is f (, ) = = ( s t s t s s (s s )) a + a ( + s t s t s s (s s )) s t s t s s (s s ). = 0. () Notice that the coefficients a and a have the same denominator. This denominator is 0 when s = 0, s = 0, or s = s. In these cases, there is no parabola through the rotated points. The following lemma shows that these conditions match our intuition about how rotations b θ, θ and θ 3, as pictured in Figure 3, do not ield parabolas. Lemma: There is no parabola through the points (, ), (, ) and (0, 0) with ais of smmetr parallel to = m i for i = to 3 where m =, m =, or m 3 =. () Proof: The equation of the parabola in () is undefined when s s (s s )= 0, which is equivalent to s = 0, s = 0 or s = s. These conditions reduce to tan θ = /, tan θ = / or ( ) sin θ = ( ) cos θ, respectivel. The last one can be rewritten as tan θ =., D5 / C C: C 5 D

5 ALL PARABOLAS THROUGH THREE NON-COLLINEAR POINTS 07 Because tan θ = /m, the slopes for which a parabola does not eist are those in () with as in Figure 3. θ i To get the final parabola, as in the description of the eample, the parabola from () is rotated counterclockwise b θ, using the rotation matri R θ. As before, the parabola P is the set of (, ) for which f ( cos θ sin θ, sin θ + cos θ) = 0. Hence, the parabola through the three initial points (, ), (, ) and (0, 0) is given b the equation a (cos θ sin θ) + a (cos θ sin θ) cos θ sin θ = 0. B substituting in the values for a and a, writing the parabola in standard form, and using a considerable amount of routine algebra and the identit cos θ + sin θ =, we obtain the following theorem. Theorem: The parabola through (, ), (, ) and (0, 0) with ais of smmetr parallel to = m has equation A + B + C + D + E + F = 0 (3) where θ = tan (/m) and A = B = ( ) cos θ ( ) ( cos θ sin θ) C = ( ) sin θ D = ( ) cos θ +( )( cos θ sin θ) +( ) sin θ E = ( ) cos θ +( )( cos θ sin θ) +( ) sin θ F = Focus, directri and verte Because a parabola is the locus of points equidistant from a point (the focus) and a line (the directri), we determine the focus, directri, and verte of the parabola with ais of smmetr of angle θ through (, ), (, ) and (0, 0). First we find the focus, directri, and verte of the equation of the parabola = a + a from () and then rotate the results b θ. The equation of a parabola ma be represented as 4p ( k) = ( h), which has focus (h, k + p), directri = k p, and verte (h, k). Completing the square in the equation = a + a ields + ( a 4a ) = a ( + a a )., D5 / C C: C 5 D

6 08 THE MATHEMATICAL GAZETTE This implies that 4p = /a. Because it is cumbersome to give the focus, directri, and verte for the parabolic function in terms of θ, we leave them in terms of the in Table. a i Focus Directri Verte ( a, a ) = a + ( a, a ) a 4a 4a a 4a TABLE : The focus, directri, and verte for the equation of the parabola = a + a To get the focus, directri, and verte of the final parabola, we need onl rotate b the focus, directri, and verte from Table counterclockwise b θ. The results appear in Table. Focus ( cosθa a + sinθ( a ) 4a, sinθa a + cosθ( a ) 4a ) Directri cos θ + sin θ = a + 4a ( Verte cosθa sinθ, sinθa cosθa ) a 4a a 4a TABLE : The focus, directri, and verte for the parabola through (, )(,, ) and (0, 0) 4. An application: the widest parabola Our presentation of the infinitel man parabolas through (, ), (, ) and (0, 0) makes it straightforward to find the widest parabola through the points. Because the final parabola is just a rotation of the parabola = a + a, the widest parabola is the rotation of the parabola from () with the minimal absolute value of a (the coefficient of ). Recall s t s t that a =. With a little algebra, one notices that the s s (s s ) numerator of a is a constant in θ. In particular, s t s t =. Taking the derivative of with respect to θ gives a (θ) = a [ (s s s ) s +(s s s ) s ] ( ) s s (s s ), where s i and t j are functions of θ. This is cumbersome to evaluate in general, but is possible for specific cases. Because a ma be negative or positive, the minimal absolute value ma be a local minimum when a is positive or a local maimum when a is negative. Returning to our motivating eample where (, ) = (, 3) and (, ) = (, ), then a (θ) = 0, when [ (s s s ) s +(s s s ) s ] ( ) = 0,, D5 / C C: C 5 D

7 ALL PARABOLAS THROUGH THREE NON-COLLINEAR POINTS 09 which reduces to 5 [ (cos θ sin θ) 39 (cos 3θ + sin 3θ] = 0. 4 From this, a takes on a local minimum at θ and this is the global minimum of a. There are two other critical points that are local maima that occur when a is negative. Each of these local maima is a candidate for the widest parabola as a is at a local minimum. Substituting θ into (3), the widest parabola through our three initial points is approimatel = 0, as shown in Figure 4. 4 (,3) (, ) (0,0) FIGURE 4: The widest parabola through the points (, 3), (, ) and (0, 0) is approimatel = 0 In contrast, there is no narrowest parabola through the points (, ), (, ) and (0, 0). This follows because a approaches positive or negative infinit as θ approaches θ i for i = to 3 (because the denominator of a approaches 0) and the greater a, the narrower the parabola. The relationship between a and θ also eplains wh there are three values of θ to consider for the widest parabola. The idea is that the θ i divide the interval ( π /, π /] (where π / and π / are identified, so that the interval is reall a circle) into three regions and a takes on a local minimum on each region. Acknowledgement: We thank an anonmous referee whose advice improved the presentation considerabl and who noticed an error. 0.07/mag.08.5 STANLEY R. HUDDY Fairleigh Dickinson Universit, Teaneck, NJ, USA srh@fdu.edu MICHAEL A. JONES Mathematical Reviews, Ann Arbor, MI, USA maj@ams.org, D5 / C C: C 5 D