Mathematics Dictionary Dr. K. G. Shih

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Conic Section : Hyperbola

Questions

Read Symbol Definition Q01 | - Defintion of Locus of a hyperbola in rectangular coordinates Q02 | - Equations of hyperbola in rectangular coordinates Q03 | - Equations or hyperbola in polar form Q04 | - Equations of hyperbola in parametric form Q05 | - Find hyperbola if focus F(2,0), vertex U(0,0) and P(2,6) are given Q06 | - Hyperbola : (x-2)^2/3^2 - (y+3)^2/4^2 = 1, find coordinates of foci Q07 | - Hyperbola : Focus F(0,0), directrix x=-D. Find equation if e=2 Q08 | - Convert x^2/4^2 - y^2/3^2 = 1 to polar form Q09 | - Convert R = 6/(1-2*cos(A)) to (x-h)^2/a^2 - y^2/b^2 = 1 Q10 | - Example : Change F(x,y) = 0 to standard form Q11 | - Formula Q12 | - Reference Q13 | - Prove the locus of hyperbola is x^2/a^2 - y^2/b^2 = 1 Q14 | - Convert (x+f)^2/a^2 - y^2/b^2 = 1 to polar form Q15 | - Draw tangent to hyperbola by law of reflextion Q16 | - Prove that D*e = (f-a)*(1+e) Q17 | - Elliminate x*y terms in F(x,y) Q18 | - Compare polar form with standard rectangular form Q19 | - Compare x^2/a^2 - y^2/b^2 = 1 with x^2/a2 - y^2/b^2 = -1

Answers

Q1. Defintion of locus of a hyperbola in rectangular coordinates A1. Defintion Figure 1 Two fixed points F and G. A moving point P(x,y). When P moves so that |PF| - |PG| = constant = 2*a, what is the locus ? Answer : The locus of point P is a hyperbola. Defintion Where F and G are the foci. The value a is the major semi-axis. The foci F and G are on the principal axis. The equation is The center of the hyperbola is C(h,k). (x-h)^2/a^2 - (y-k)^2/b^2 = +1 if the princial axis in x-axis direction. (x-h)^2/a^2 - (y-k)^2/b^2 = -1 if the princial axis in y-axis direction. Focal length : f = CF = CG = focal length = Sqr(a^2+b^2). Vertices on principal axis : CU = a and CV = a. Distance between focus and vertex is FU = f - a. Go to Begin Q2. Equations of hyperbola A2. Answers [Standard Form] Equation : (x-h)^2/a^2 - (y-k)^2/b^2 = 1. Where C(h,k) is the center. The values a and b are semi-axis. The principal axis is along x-axis The principal axis is y = k. The foci F and G are on principal axis. The vertice U and V are on pricipal axis. Equation : (x-h)^2/a^2 - (y-k)^2/b^2 = -1. Where C(h,k) is the center. The values a and b are semi-axis. The principal axis is along y-axis The principal axis is x = h. The foci F and G are on principal axis. The vertice U and V are on pricipal axis. The focal length is CF or CG which is Sqr(a^2+b^2). The vertice The end points alnog principal axis is U and V. The a = CU = CV. FU = CF - CU = f - a. [Implicit Form without x*y] Equation : F(x,y) = A*x^2 + C*y^2 + D*x + E*y + F = 0. Hyperbola : A and C must have different signs. Princial axis is parallel to x-axis or y-axis. Foci are on princial axis. Find foci, a, b and f of hyperbola. Change to standard form by using completing the square. The standard form is (x-h)^2/a^2 - (y-k)^2/b^2 = 1. Hence we can find the center (h,k) and semi-axis a and b. Hence we can find f using f = Sqr(a^2+b^2). [Implicit Form with term x*y] Equation : F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0. B^2 - 4*A*C > 0 if it is a hyperbola. The pricipal axis is y = m*x + n and not parallel to x-axis or y-axis. How to find semi-axis and foci ? We must elliminate x*y by rotating an angle. The new coefficients A' C' D' E' F' can be obtained by using B' = 0. Then we can find a, b, f and coordinates of foci. Detailed method is given in ZE.txt or ZC.txt of MD2002. [Example] Study (x+2)^2/4^2 - (y-3)^2/3^2 = 1 Pricipal axis is y = 3. Center at (-2,3), a=4 and b=3. Focal length f = Sqr(a^2+b^2) = Sqr(5^2+3^2) = 5. Foci at (-7,3) and (3,3). [Example] Study (x+2)^2/3^2 - (y-3)^2/4^2 = -1 Pricipal axis is x = -2. Center at (-2,3), a=4 and b=3. Focal length f = Sqr(a^2+b^2) = Sqr(4^2+3^2) = 5. Foci at (-2,8) and (-2,-2). [Example] Study 9*x^2 - 25*y^2 - 54*x + 100*y - 44 = 0 9*(x^2 - 6*x + 9) - 81 - 25*(y^2 - 4*y + 4) - 100 - 44 = 0. 9*(x-3)^2 - 25*(y-2)^2 = 225. (x-3)^2/5^2 - (y+2)^2/3^2 = 1. Hence it is a hyperbola. Pricipal axis is y = -2. Center at (3,-2), a=5 and b=3. Focal length f = Sqr(a^2+b^2) = Sqr(5^2+3^2) = Sqr(34). Foci at x = h - f, y = k and x = h + f, y = k. Go to Begin Q3. Equations in Polar Form Figure 2 Locus in polar coordinate [Function 1] Function : R = D*e/(1-e*sin(A)). Origin is at F and directrix is y = -D. D is focus to directrix line and e is eccentricy greater than 1. A is angle making with the x-axis. Foci are on x-axis. [Function 2] Function : R = D*e/(1+e*sin(A)). Origin is at F and directrix is y = D. D is focus to directrix line and e is eccentricy greater than tha 1. A is angle making with x-axis. Foci are on y-axis and origin is on top focus. [Function 3] Function : R = D*e/(1-e*cos(A)). Origin is at F and directrix is x = -D. D is focus to directrix line and e is eccentricy greater than 1. A is angle making with x-axis. Foci are on x-axis and origin is on left focus. [Function 4] Function : R = D*e/(1+e*cos(A)). Origin is at F and directrix is x = D. D is focus to directrix line and e is eccentricy greater than 1. A is angle making with x-axis. Foci are on x-axis and origin is on right focus. [Example] Relation of R=D*e/(1-e*cos(A)) and (x-f)^2/a^2 - y^2/b^2 = 1. R = D*e/(1-e*cos(A)) : Origin at F and e greater than 1. (x-f)^2/a^2 - y^2/b^2 = 1 and let it have same focus F. When A = 180 degrees then cos(180) = -1. Hence R = D*e/(1+e). From diagram we know R = FU = f-a when A = 180 degrees. Hence D*e = (f-a)*(1+e). [Example] Prove that R = D*e/(1-e*cos(A)) is hyperbola Draw vertical line as directrix. Draw princial axis perpendicular to the directrix. Let F on princial axis as origin (0,0). Draw a point P(x,y). Let PQ = distance from P to Q where Q is on directrix. Locus of P is hyperbola if PF/PQ = e and greater than 1. Where PF = R and PQ = D+x. D = distance from F to directrix. Prove that R = D*e/(1-e*cos(A)). R/PQ = e. R = e*PQ = e*(D+x). Since x = r*cos(A). Hence R = e*(D+e*R*cos(A)) Hence R = e*D/(1-e*cos(A)) Go to Begin Q4. Equation in parametric form A4. Answer x = h + a*sec(t). y = k + b*tan(t). (h,k) are center. The values of a and b are semi-axis. It is hyperbola since (x-h)^2/a^2 - (y-k)^2/b^2 = 1. Go to Begin Q5. Example : Find equation hyperbola if 3 points are given Give focus F(2,0) and vertix U(0,0). A point is P(2,6) on the hyperbola. Find equation of the hyperbola. Method 1 Construction Draw a horizon line as principal axis (x-axis). Draw U(0,0) as origin and vertex of the hyperbola. Draw focus F(2,0) on x axis. Draw point P(2,6) Find the equation of the parabola. Let the center of parabola be C(h,0). Then the equation is (x-h)^2/a^2 - y^2/b^2 = 1. Find h, f, a and b CU = a, CF = f and h = -a. Then UF = CF - CU = f - a = 2 or f = 2 + a. It is known that b^2 = f^2 - a^2 = (2+a)^2 - a^2. Or b^2 = 4*(1+a). Substitue h, b^2 into above we have (x+a)^2/a^2 - y^2/(4+4*a) = 1. Substitue x=2 and y=6 into equation we have (2+a)/a^2 - 36/(4+4*a) = 1. Solve for a and we get a = 2. Hence center is at x = -2. Then f = 4. And b^2 = f^2 - a^2 = 4^2 - 2^2 = 12 Hence equation is (x+2)^2/4 - y^2/12 = 1. Method 2 : Polar equation Draw rectangular coordinates system. Translate above points to left by 2 units. That is F(0,0), U(-2,0) and P(0,6). Let Q be point on diretrix and PQ be perpendicular to directrix. Then PF = 6 and UF = 2. Then polar equation is R = D*e/(1-e*cos(A)). Find D*e. If A = 90 degrees, cos(A) = 0. R = PF = 6. Then 6 = D*e/(1+0) or D*e = 6. Find e If A = 180 degrees, cos(A) = -1. R = UF = 2. Then 2 = D*e/(1+e) or 2*(1+e) = D*e = 6. Hence e = 2. Hence R = 6/(1-2*cos(A)) Method 3 : Conver polar equation to (x-h)^2/a^2 - (y-k)^2/b^2 = 1 Find h, a, k, b. From method 2 we have e = 2 and D = 3. Also center C(h,k) where h = -a and k = 0. CF = f and CU = a. Find a and f Since e = f/a = 2 or f = 2*a. When A = 180 degrees, R = UF = 2 = CF - CU = f - a. Hence f = 2 + a or 2*a = 2 + a. Hence a = 2 and f = 4. Find b^2 Since b^2 = f^2 - a^2 = 4^2 - 2^2 = 16 - 4 = 12. Substitute a, b, h and k into equation. Hence equation is (x+2)^2/4 - y^2/12 = 1. Go to Begin Q6. Example : If (x-2)^2/3^2 - (y+3)^2/4^2 = 1, find coordinates of foci Principal axis is y = -3. The foci are on principal axis. The center is at C(2,-3). Focal length f = Sqr(3^2 + 4^2) = 5. Coordinate of focus is at F(2,-8). Coordinate of other focus is at G(2,2). Go to Begin Q07. Hyperbola : Focus F(0,0), directrix x = -3. Find equation if e = 2 [Method 1] Polat form is R = D*e/(1-e*cos(A)). D = distance of F to directrix = 3. The eccentricity is e = 2. Hence the polar equation is R = 6/(1-2*cos(A)). [Method 2] The rectangular form is (x-h)^2/a^2 - y^2/b^2 = 1 Since F is the origin, and the principal axis is y = 0. Hence the center is C(h,0). Let U be the vertex and then CU = a and CF = f. Also h = -f. Find relation of f and a. UF = f - a. When A = 180, R = UF = D*e/(1+e). Hence D*e = (1+e)*(f-a). Hence 6 = (1+2)*(f-a) or f-a = 2. e = f/a and f = e*a = 2*a. Hence a = 2 and f = 4 Find b Since f = Sqr(a^2+b^2) or b^2 = f^2 - a^2. b^2 = 4^2 - 2^2 = 12. The equation is (x+4)^2/2^2 - y^2/12 = 1. Go to Begin Q8. Example : Convert x^2/4^2 - y^2/3^2 = 1 to polar form The polar form is R = D*e/(1-e*cos(A)) F is origin Directrix is x = -D. When A=180 degrees, R = f-a. Hence f - a = D*e/(1+e) or D*e = (f-a)*(1+e). x^2/4^2 - y^2/3^2 = 1 Left focus is at (-5,0). Focal length f = Sqr(4^2+3^2) = 5. The eccentricity e = f/a = 5/4 = 1.25. D*e = (a-f)*(1+e) = (5-4)*(1+1.25) = 2.25. Hence required polar function is R = 2.25/(1-1.25*cos(A)) Go to Begin Q9. Example : Convert R = 2.25/(1-1.25*cos(A) to (x-h)^2/a^2 - y^2/b^2 = 1 R = D*e/(1-e*cos(A)) D*e = 2.25 and e = 1.25. When A = 180, R = f-a = D*e/(1+e) Hence D*e = (f-a)*(1+e) = 2.25 e = f/a = 1.25. Find a Substitute f = 1.25*a into D*e. Hence (1.25*a-a)*(1+1.25) = 2.25. Hence 0.25*a = 1 and hence a = 4. Find b f = 5 and a = 4. Since f = Sqr(a^2 + b^2) = 5 and hence b = 3. The required equation is x^2/4^2 - y^2/3^2 = 1. Go to Begin Q10. Example : 9*x^2 - 16*y^2 - 18*x - 64*y - 71 = 0, find a, b ,f Change it to standard form by completing the square. 9*(x^2 - 2x + 1 - 1) - 16*(y^2 + 4*y + 4 - 4) - 71 = 0. 9*(x-1)^2 - 9 + 25*(y+2)^2 - 64 - 71 = 0. 9*(x-1)^2 + 16*(y+2)^2 = 144. Divide both sides by 144 and we have : (x-1)^2/4^2 - (y+2)^2/3^2 = 1. Hence a = 4 and b = 3. Then f = Sqr(a^2+b^2) = 5. Principal axis is y = -2. This is a hyperbola because (B^2 - 4*A*C) = 0 + 4*9*25 = positive. Go to Begin Q11. Formula : (x-h)^2/a^2 - (y-k)^2/b^2 = 1 Definiton from construction Principal axis is y = k. Vertice on principal axis are at U and V. Foci on principal axis are at F and G. Center is at (h,k). CU = CV = a = Semi-axis on principal axis. Focal length f = Sqr(a^2 + b^2). e = f/a. f = CF = CG Efficiency e = f/a. Locus in rectangular system |PF| - |PG| = 2*a. F and G are foci. P is moving point. Equation is (x-h)^2/a^2 - (y-k)^2/b^2 = 1 Locus in polar system R/PQ = e. P is moving point. PQ is distance from P to Q on directrix. F is origin and PF = R. Equation is R = D*e/(1-e*cos(A)) and e greater than 1. Relation between polar and rectangular : R = D*e/(1-e*cos(A)). R = UF = (f-a) when A = 180 degrees. Use polar formula we have D*e = (f-a)*(1+e). D is the distance from focus F to directrix. Slope of point on hyperbola dy/dx = (x*b^2)/(y*a^2). Asymptotes y - k = +b*(x-h)/b y - k = -b*(x-h)/b Go to Begin Q12. References : Sketch (x-h)^2/a^2 - (y-h)^2/b^2 = 1 ..... MD2002 ZM40 04 Sketch R = D*e/(1-e*sin(A)) .............. MD2002 ZM40 08 Sketch F(x,y) = 0 ........................ MD2002 Conic Section ............................ MD2002 ZM06 and ZM40 Elliminate x*y terms in F(x,y)=0 ......... MD2002 ZM34 12 See keywords hyperbola in Index File See contents of ZH.txt in Contents by chapres Go to Begin Q13. Example : Prove that locus of hyperbola is x^2/a^2 - y^2/b^2 = 1 Let the fixed points be F(x,-f) and G(x,f). Moving point is P(x,y). Since |PF} - |PG| = 2*a where a is the major semi-axis. Sqr((x+f)^2 + y^2) - Sqr((x-f)^2 + y^2) = 2*a. Square bothe sides we have : (x+f)^2+ y^2+ (x-f)^2+ y^2- 2*Sqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2)=4*a^2. x^2+2*x*f+f^2+y^2+x^2-2*x*f+f^2+f^2-4*a^2=2*Aqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2). or 2*x^2+ 2*y^2+ 2*f^2- 4*a^2 = 2*Sqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2). or (x^2+ y^2)+ (f^2- 2*a^2) = Sqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2). Square both sides again : (x^2+y^2)^2+2*(x^2+y^2)+(f^2-2*a^2)^2 = (x+f)^2+y^2)*(x-f)^2+y^2). x^4+2*x^2*y^2+y^4+ 2*f^2*(x^2+y^2)- 4*a^2*(x^2+y^2)+ f^4-4*f^2*a^2 +4*a^2. = (x+f)^2*(x-f)^2 + y^2*(x-f)^2 + y^2*(x+f) +y^4. = x^4-2*x^2*y^2+f^4+x^2*y^2-2*x*f*y^2+y^2*f^2+x^2*y^2+2*x*f*y^2+y^2*f^2+y^4. Simplify above equation : 4*x^2*f^2 - 4*a^2*x^2 + 4*a^2*y^2 = 4*a^2*f^2 - 4*a^4. -(4*a^2-4*f^2)*x^2 + 4*a^2*y^2 = 4*a^2*(a^2-b^2) - 4*a^4 . -(a^2-f^2)*x^2 + a^2*y^2 = -a^2*b*2. b^2*x^2 - a^2*y^2 = a^2*b^2. Hence equation of locus is x^2/a^2 - y^2/b^2 = 1. By translation : (x-h)^2/a^2 + (y-k)^2/b^2 = 1. Go to Begin Q14. Example : Convert (x+f)^2/a^2 - y^2/b^2 = 1 to polar form The focal point is at F(x,-f) for polar form R = D*e/(1-e*cos(A)). Remove denominator : b^2*(x+f)^2 - a^2*y^2 = a^2*b*2. b^2*x^2 + b^2*x*f + b^2*f^2 - a^2*y^2 = a^2*b^2. Since b^2 = f^2 - a^2. Hence b^2*x^2 - a^2*y^2 + 2*b^2*x*f = a^2*b^2 - b^2*f^2. (f^2-a^2)*x^2 - a^2*y*2 + 2*b^2*x*f = b^2*(f^2-a^2) Since x = R*cos(A) and y = R*sin(A) and R^2 = x^2 + y^2. Simplify above equation : -R^2*a^2 + f^2*x^2 + 2*b^2*x*f = b^4. -R^2*a^2 + (f^2*x^2 + 2*b^2*x*f + b^4) + b^4 = b^4. R^2*a^2 - (f*x + b^2)^2 = 0. R*a = -(f*x + b^2). R*a = +(f*x + b^2). R*a - R*f*cos(A) = b^2 R*(a - f*cos(A)) = b^2. R = b^2/(a - f*cos(A)) After elliminting f by f=e/a we have : R = b^2/(a - e*cos(A)/a). Hence R = (b^2/a)/(1 - e*cos(A)). If A = 180 we have R = (b^2/a)/(1+e) or b^2/a = R*(1+e). b^2/a = (f^2 - a^2)/a = (f-a)*(a+f)/a = (f-a)*(1+e) = D*e. Hence R = D*e/(1-e*cos(A)). Other method : See Q08 Go to Begin Q15. Example : Draw tangent to hyperbola using reflection Figure 3 Tangent by reflection Draw an hyperbola with F(x,-f) and G(x,f). Draw a point P(x,y) on the hyperbola. Draw a bisector of angle FPG. Draw a line perpendicular the bisector and passing P. By the law of reflection, This line is the requred tangent. Go to Begin Q16. Example : Relation of D*e = (f-a)*(1+e) Draw an hyperbola with F(x,-f) and G(x,f) on x-axis. Let the vertex U be between directrix and F. When A = 180 then R = D*e/(1+e) = FU = f-a. Hence D*e = (f-a)*(1+e). Where e = f/a and f = Sqr(a^2-b^2). Go to Begin Q17. Elliminate x*y terms in F(x,y) Equation : A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0. Use rotation formula to change above equation as Equation : A'*x^2 + B'*x*y + C'*y^2 + D'*x + E'*y + F' = 0. Let B' = 0 and then we can find the rotation angle. Then we can find A', C', D', E', F'. Then we can find a, b, f, and the cneter (h,k) and principal axis. Examples in Conic Sections Go to Begin Q19. Compare x^2/a^ - y^2/b^2 = 1 and x^2/a^2 - y^2/b^2 = -1 x^2/a^ - y^2/b^2 = 1 Principal axis is y = 0. Foci are at (-f,0) and (f,0). Vertice on principal axis are (-a,0) and (a,0). x^2/a^ - y^2/b^2 = -1 Principal axis is x = 0. Foci are at (0,-f) and (0,f). Vertice on principal axis are (0,-a) and (0,a). Go to Begin

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