Notes On Standard Equations of Hyperbola - CBSE Class 11 Maths
A hyperbola is the set of all that points in a plane such that the difference of their distances from two fixed points in the plane is constant. P1F2 - P1F1 = P2F2 - P2F1 = P3F1 - P3F2 Foci, vertices, centre, transverse axis and conjugate axis of a hyperbola are as shown in the figure. In a hyperbola, b2 = c2 - a2, Where c = Distance of a focus from the centre a = Length of semi-transverse axis b = Length of semi-conjugate axis The difference between the distances of any point on a hyperbola from its foci is equal to the length of the transverse axis of the hyperbola. P1F2 - P1F1 = P2F2 - P2F1 = P3F1 - P3F2 = 2a Eccentricity is a measure of the deviation of a conic section from being a circle. The eccentricity of a hyperbola is denoted by E, and is equal to the ratio of the distance of a focus from its centre and the length of its semi-transverse axis. Eccentricity of a hyperbola (e) = c/a Þ c = ea Thus, the distance of a focus from the centre of a hyperbola is equal to the product of its eccentricity and the length of its semi-transverse axis. Standard Equations of a Hyperbola Consider a hyperbola with its centre at the origin and its transverse axis along the X- or the Y-axis, and then there are two distinct possibilities. In the first case, we can have a hyperbola with its centre lying on the origin and its transverse axis along the X-axis. In this case, the foci of the hyperbola lie on the X-axis. In the second case, we can have a hyperbola with its centre lying on the origin and its transverse axis along the Y-axis. In this case, the foci of the hyperbola lie on the Y-axis. Case (i) Equation of the hyperbola: Consider a point P on the hyperbola with the coordinates X, Y. Then, by definition, the difference of the distances PF1 and PF2 is constant and equal to 2A. PF1 - PF2 = 2a ……(1) The distance between the points (x1, y1) and (x2, y2) = √(x2 - x1)2 + (y2 - y1)2 Using distance formula, PF1 = √{(x - (-c)}2 + (y - 0)2 ⇒ PF1 = √(x + c)2 + y2 ……(2) Using distance formula, PF2 = √(x - c)}2 + (y - 0)2 ⇒ PF2 = √(x - c)2 + y2 ……(3) From equations (1), (2) and (3), we get √(x + c)2 + y2 - √(x - c)2 + y2 = 2a ⇒ √(x + c)2 + y2 = 2a + √(x - c)2 + y2 Squaring both sides, (x + c)2 + y2 = 4a2 + 4a√(x - c)2 + y2 + (x - c)2 + y2 ⇒ 4a√(x - c)2 + y2 = (x + c)2 - (x - c)2 - 4a2 ⇒ 4a√(x - c)2 + y2 = x2 + c2 + 2xc - x2 - c2 + 2xc - 4a2 ⇒ 4a√(x - c)2 + y2 = 4xc - 4a2 ⇒ a√(x - c)2 + y2 = xc - a2 ⇒ √(x - c)2 + y2 = c/a x - a Squaring both sides, (x - c)2 + y2 = a2 + c2/a2 x - 2cx ⇒x2 + c2 - 2cx + y2 = a2 + c2/a2 x2 - 2cx ⇒x2 + c2 + y2 = a2 + c2/a2 x2 ⇒x2 - c2/a2 x2 + y2 = a2 - c2 ⇒(1 - c2/a2)x2 + y2 = a2 - c2 ⇒((a2 - c2)/a2) x2 + y2 = a2 - c2 ……(4) ⇒( - b2/a2) x2 + y2 = -b2 (Since b2 = c2 - a2) Multiplying both sides by -1/b2, x2/a2 - y2/b2 = 1 ……(5) Therefore, equation x2/a2 - y2/b2 = 1 represents the given hyperbola if the coordinates of point P as derived from this equation satisfy the geometrical condition (PF1 - PF2) = 2a. From equation (5), y2 = b2 (x2/a2 - 1) From equation (2), PF1 = √(x + c)2 + y2 ⇒ PF1 = √(x + c)2 + b2 (x2/a2 - 1) = √ x2 + c2 + 2cx + b2x2/a2 - b2 = √ x2(1+ b2/a2 + 2cx + c2 - b2 = √ x2((a2+b2)/a2)+ 2cx + c2 - b2 = √c2/a2 .x2+ 2cx + a2 (Since b2 = c2 - a2) = √(a + cx/a)2 ⇒ PF1 = a + cx/a From equation (3), PF2 = √(x - c)2 + y2 ⇒ PF1 = √(x - c)2 + b2 (x2/a2 - 1) = √ x2 + c2 - 2cx + b2x2/a2 - b2 = √ x2(1+ b2/a2) - 2cx + c2 - b2 = √ x2( (a2+ b2)/a2) - 2cx + c2 - b2 = √c2/a2 . x2 - 2cx + a2 (Since b2 = c2 - a2) = √( c/a x - a)2 ⇒ PF2 = c/a x - a ⇒ PF1 - PF2 = a + c/a . x - c/a . x + a = 2a Case I: The equation, x2/a2 - y2/b2 = 1, is the standard equation for a hyperbola with its centre at the origin and the transverse axis lying along the X-axis. Case II: Similarly, the standard equation for a hyperbola with its centre at the origin and the transverse axis along the Y-axis is y2/a2 - x2/b2 = 1. These two equations are called the standard equations of a hyperbola having its centre at the origin and the transverse axis along the X- or Y-axis. If the coefficient of x2 is positive, then the transverse axis lies along the X-axis. If the coefficient of y2 is positive, then the transverse axis lies along the Y-axis. Note that both the standard equations of a hyperbola contain even powers of x and y. A hyperbola with its centre lying on the origin and the transverse axis along either the X- or the Y-axis is symmetrical about both the coordinate axes. ⇒ If (x, y) lies on the hyperbola, then (-x, y), (x, -y) and (-x, -y) also lie on it. If a = b, then the hyperbola is called an equilateral hyperbola. Case I: Centre (0, 0), transverse axis along the X-axis. x2/a2 - y2/b2 = 1 ⇒ x2/a2  = 1 + y2/b2 ⇒ |x/a | ≥ 1 ⇒x ≤ -a or x ≥ a ⇒ No point on the curve of the hyperbola lies between the lines x = -a and x = a. Case II: Centre (0, 0), transverse axis along the Y-axis.     y2/a2 - x2/b2 = 1 ⇒ y2/a2 = 1 + x2/b2 ⇒ |y/a |≥ 1 ⇒ y ≤ -a or y ≥ a ⇒ No point on the curve of the hyperbola lies between the lines y = -a and y = a. Expression for Length of Latus Rectum A line segment passing through a focus and perpendicular to the transverse axis, and with its end points lying on the curve of a hyperbola, is called its latus rectum. Since a hyperbola has two foci, it has two latus recta. Consider a hyperbola with its centre at the origin and the transverse axis along the X-axis and latus recta AB and CD. Let AF2 = l ⇒ Coordinates of A are (c, l), where c = Distance of focus F2 from centre O Equation of the given hyperbola: x2/a2 - y2/b2 = 1 Sine A (c, l) lies on x2/a2 - y2/b2 = 1 ⇒ c2/a2 - l2/b2 = 1 ⇒ l2 = b2(c2/a2 - 1) ⇒ l2 = b2((c2-a2) / a2) ⇒ l2 = b2( b2/ a2) (Since b2 = c2 - a2) ⇒ l2 = b4/ a2 ⇒ l = b2/a AF2 = F2B = b2/a ⇒ AB = AF2 + F2B = 2b2/a Length of latus rectum of a hyperbola = 2b2/a, Where, a = Length of semi-transverse axis b = Length of semi-conjugate axis.

#### Summary

A hyperbola is the set of all that points in a plane such that the difference of their distances from two fixed points in the plane is constant. P1F2 - P1F1 = P2F2 - P2F1 = P3F1 - P3F2 Foci, vertices, centre, transverse axis and conjugate axis of a hyperbola are as shown in the figure. In a hyperbola, b2 = c2 - a2, Where c = Distance of a focus from the centre a = Length of semi-transverse axis b = Length of semi-conjugate axis The difference between the distances of any point on a hyperbola from its foci is equal to the length of the transverse axis of the hyperbola. P1F2 - P1F1 = P2F2 - P2F1 = P3F1 - P3F2 = 2a Eccentricity is a measure of the deviation of a conic section from being a circle. The eccentricity of a hyperbola is denoted by E, and is equal to the ratio of the distance of a focus from its centre and the length of its semi-transverse axis. Eccentricity of a hyperbola (e) = c/a Þ c = ea Thus, the distance of a focus from the centre of a hyperbola is equal to the product of its eccentricity and the length of its semi-transverse axis. Standard Equations of a Hyperbola Consider a hyperbola with its centre at the origin and its transverse axis along the X- or the Y-axis, and then there are two distinct possibilities. In the first case, we can have a hyperbola with its centre lying on the origin and its transverse axis along the X-axis. In this case, the foci of the hyperbola lie on the X-axis. In the second case, we can have a hyperbola with its centre lying on the origin and its transverse axis along the Y-axis. In this case, the foci of the hyperbola lie on the Y-axis. Case (i) Equation of the hyperbola: Consider a point P on the hyperbola with the coordinates X, Y. Then, by definition, the difference of the distances PF1 and PF2 is constant and equal to 2A. PF1 - PF2 = 2a ……(1) The distance between the points (x1, y1) and (x2, y2) = √(x2 - x1)2 + (y2 - y1)2 Using distance formula, PF1 = √{(x - (-c)}2 + (y - 0)2 ⇒ PF1 = √(x + c)2 + y2 ……(2) Using distance formula, PF2 = √(x - c)}2 + (y - 0)2 ⇒ PF2 = √(x - c)2 + y2 ……(3) From equations (1), (2) and (3), we get √(x + c)2 + y2 - √(x - c)2 + y2 = 2a ⇒ √(x + c)2 + y2 = 2a + √(x - c)2 + y2 Squaring both sides, (x + c)2 + y2 = 4a2 + 4a√(x - c)2 + y2 + (x - c)2 + y2 ⇒ 4a√(x - c)2 + y2 = (x + c)2 - (x - c)2 - 4a2 ⇒ 4a√(x - c)2 + y2 = x2 + c2 + 2xc - x2 - c2 + 2xc - 4a2 ⇒ 4a√(x - c)2 + y2 = 4xc - 4a2 ⇒ a√(x - c)2 + y2 = xc - a2 ⇒ √(x - c)2 + y2 = c/a x - a Squaring both sides, (x - c)2 + y2 = a2 + c2/a2 x - 2cx ⇒x2 + c2 - 2cx + y2 = a2 + c2/a2 x2 - 2cx ⇒x2 + c2 + y2 = a2 + c2/a2 x2 ⇒x2 - c2/a2 x2 + y2 = a2 - c2 ⇒(1 - c2/a2)x2 + y2 = a2 - c2 ⇒((a2 - c2)/a2) x2 + y2 = a2 - c2 ……(4) ⇒( - b2/a2) x2 + y2 = -b2 (Since b2 = c2 - a2) Multiplying both sides by -1/b2, x2/a2 - y2/b2 = 1 ……(5) Therefore, equation x2/a2 - y2/b2 = 1 represents the given hyperbola if the coordinates of point P as derived from this equation satisfy the geometrical condition (PF1 - PF2) = 2a. From equation (5), y2 = b2 (x2/a2 - 1) From equation (2), PF1 = √(x + c)2 + y2 ⇒ PF1 = √(x + c)2 + b2 (x2/a2 - 1) = √ x2 + c2 + 2cx + b2x2/a2 - b2 = √ x2(1+ b2/a2 + 2cx + c2 - b2 = √ x2((a2+b2)/a2)+ 2cx + c2 - b2 = √c2/a2 .x2+ 2cx + a2 (Since b2 = c2 - a2) = √(a + cx/a)2 ⇒ PF1 = a + cx/a From equation (3), PF2 = √(x - c)2 + y2 ⇒ PF1 = √(x - c)2 + b2 (x2/a2 - 1) = √ x2 + c2 - 2cx + b2x2/a2 - b2 = √ x2(1+ b2/a2) - 2cx + c2 - b2 = √ x2( (a2+ b2)/a2) - 2cx + c2 - b2 = √c2/a2 . x2 - 2cx + a2 (Since b2 = c2 - a2) = √( c/a x - a)2 ⇒ PF2 = c/a x - a ⇒ PF1 - PF2 = a + c/a . x - c/a . x + a = 2a Case I: The equation, x2/a2 - y2/b2 = 1, is the standard equation for a hyperbola with its centre at the origin and the transverse axis lying along the X-axis. Case II: Similarly, the standard equation for a hyperbola with its centre at the origin and the transverse axis along the Y-axis is y2/a2 - x2/b2 = 1. These two equations are called the standard equations of a hyperbola having its centre at the origin and the transverse axis along the X- or Y-axis. If the coefficient of x2 is positive, then the transverse axis lies along the X-axis. If the coefficient of y2 is positive, then the transverse axis lies along the Y-axis. Note that both the standard equations of a hyperbola contain even powers of x and y. A hyperbola with its centre lying on the origin and the transverse axis along either the X- or the Y-axis is symmetrical about both the coordinate axes. ⇒ If (x, y) lies on the hyperbola, then (-x, y), (x, -y) and (-x, -y) also lie on it. If a = b, then the hyperbola is called an equilateral hyperbola. Case I: Centre (0, 0), transverse axis along the X-axis. x2/a2 - y2/b2 = 1 ⇒ x2/a2  = 1 + y2/b2 ⇒ |x/a | ≥ 1 ⇒x ≤ -a or x ≥ a ⇒ No point on the curve of the hyperbola lies between the lines x = -a and x = a. Case II: Centre (0, 0), transverse axis along the Y-axis.     y2/a2 - x2/b2 = 1 ⇒ y2/a2 = 1 + x2/b2 ⇒ |y/a |≥ 1 ⇒ y ≤ -a or y ≥ a ⇒ No point on the curve of the hyperbola lies between the lines y = -a and y = a. Expression for Length of Latus Rectum A line segment passing through a focus and perpendicular to the transverse axis, and with its end points lying on the curve of a hyperbola, is called its latus rectum. Since a hyperbola has two foci, it has two latus recta. Consider a hyperbola with its centre at the origin and the transverse axis along the X-axis and latus recta AB and CD. Let AF2 = l ⇒ Coordinates of A are (c, l), where c = Distance of focus F2 from centre O Equation of the given hyperbola: x2/a2 - y2/b2 = 1 Sine A (c, l) lies on x2/a2 - y2/b2 = 1 ⇒ c2/a2 - l2/b2 = 1 ⇒ l2 = b2(c2/a2 - 1) ⇒ l2 = b2((c2-a2) / a2) ⇒ l2 = b2( b2/ a2) (Since b2 = c2 - a2) ⇒ l2 = b4/ a2 ⇒ l = b2/a AF2 = F2B = b2/a ⇒ AB = AF2 + F2B = 2b2/a Length of latus rectum of a hyperbola = 2b2/a, Where, a = Length of semi-transverse axis b = Length of semi-conjugate axis.

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