ncert books for class maths cbsc

A Note The standard equations of ellipses have centre at the origin and the major and minor axis are coordinate axes. However, the study of the ellipses with centre at any other point, and any line through the centre as major and the minor axes passing through the centre and perpendicular to major axis are beyond the scope here. From the standard equations of the ellipses (Fig11. ), we have the following observations: .

Ellipse is symmetric with respect to both the coordinate axes since if ( x , y ) is a point on the ellipse, then (– x , y ), ( x , – y ) and (– x , – y ) are also points on the ellipse. . The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry.

That is, major axis is along the x -axis if the coefficient of x has the larger denominator and it is along the y -axis if the coefficient of y has the larger denominator. MATHEMATICS . . Latus rectum Definition Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse (Fig .

). To find the length of the latus rectum of the ellipse b Let the length of AF be l . Then the coordinates of A are ( c , l ) , i.e., ( ae , l ) Since A lies on the ellipse b = , we have ae l b ⇒ l = b ( – e ) But c a – b b e – Therefore l = b , i.e., b l Since the ellipse is symmetric with respect to y -axis (of course, it is symmetric w.r.t. both the coordinate axes), AF = F B and so length of the latus rectum is b a .

Example Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse Solution Since denominator of x is larger than the denominator of y , the major Fig . CONIC SECTIONS axis is along the x -axis. Comparing the given equation with b = , we get a = and b = . Also c a – b – Therefore, the coordinates of the foci are (– , ) and ( , ), vertices are (– , ) and ( , ).

Length of the major axis is units length of the minor axis b is units and the eccentricity is and latus rectum is b a = Example Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse x + y = . Solution The given equation of the ellipse can be written in standard form as Since the denominator of y is larger than the denominator of x , the major axis is along the y -axis. Comparing the given equation with the standard equation b = , we have b = and a = . Also c = a – b = – and c e Hence the foci are ( , ) and ( , – ), vertices are ( , ) and ( , – ), length of the major axis is units, the length of the minor axis is units and the eccentricity of the ellipse is .

Example Find the equation of the ellipse whose vertices are ( ± , ) and foci are ( ± , ). Solution Since the vertices are on x -axis, the equation will be of the form b = , where a is the semi-major axis. MATHEMATICS Given that a = , c = ± . Therefore, from the relation c = a – b , we get = – b , i.e., b = Hence the equation of the ellipse is = .

Example Find the equation of the ellipse, whose length of the major axis is and foci are ( , ± ). Solution Since the foci are on y -axis, the major axis is along the y -axis. So, equation of the ellipse is of the form b = . Given that a = semi-major axis and the relation c = a – b gives = – b i.e.

, b = Therefore, the equation of the ellipse is Example Find the equation of the ellipse, with major axis along the x -axis and passing through the points ( , ) and (– , ). Solution The standard form of the ellipse is b = . Since the points ( , ) and (– , ) lie on the ellipse, we have + b ... ( ) and b = ….( ) Solving equations ( ) and ( ), we find that a = and b = Hence the required equation is CONIC SECTIONS , i.e., x + y = .

EXERCISE . In each of the Exercises to , find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. . .

= . x + y = . x + y = . x + y = In each of the following Exercises to , find the equation for the ellipse that satisfies the given conditions: .

Ends of major axis ( , ± ), ends of minor axis ( ± , ) . Length of major axis , foci ( ± , ) . Length of minor axis , foci ( , ± ). .

Foci ( ± , ), a = . b = , c = , centre at the origin; foci on the x axis. . Centre at ( , ), major axis on the y -axis and passes through the points ( , ) and ( , ).

. Major axis on the x -axis and passes through the points ( , ) and ( , ). . Hyperbola Definition A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

MATHEMATICS The term “ difference ” that is used in the definition means the distance to the farther point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola . The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis .

The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola (Fig . ). We denote the distance between the two foci by c , the distance between two vertices (the length of the transverse axis) by a and we define the quantity b as b = c – a Also b is the length of the conjugate axis (Fig . ).

To find the constant P F – P F : By taking the point P at A and B in the Fig . , we have BF – BF = AF – AF (by the definition of the hyperbola) BA +AF – BF = AB + BF – AF i.e., AF = BF So that, BF – BF = BA + AF – BF = BA = a Fig . Fig . CONIC SECTIONS .

. Eccentricity Definition Just like an ellipse, the ratio e = c a is called the eccentricity of the hyperbola . Since c ≥ a , the eccentricity is never less than one. In terms of the eccentricity, the foci are at a distance of ae from the centre.

. . Standard equation of Hyperbola The equation of a hyperbola is simplest if the centre of the hyperbola is at the origin and the foci are on the x -axis or y -axis. The two such possible orientations are shown in Fig11.

. We will derive the equation for the hyperbola shown in Fig . (a) with foci on the x -axis. Let F and F be the foci and O be the mid-point of the line segment F F .

Let O be the origin and the line through O through F be the positive x -axis and that through F as the negative x -axis. The line through O perpendicular to the x -axis be the y -axis. Let the coordinates of F be (– c , ) and F be ( c , ) (Fig . ).

Let P( x , y ) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be a. So given , PF – PF = a Fig . (a) (b) Fig . MATHEMATICS Using the distance formula, we have c – x – c i.e., c x – c Squaring both side, we get ( x + c ) + y = a + a x – c + ( x – c ) + y and on simplifying, we get cx – a = x – c On squaring again and further simplifying, we get – c – a i.e., – b (Since c – a = b ) Hence any point on the hyperbola satisfies – b = .

Conversely, let P( x , y ) satisfy the above equation with < a < c . Then y = b x – a Therefore, PF = + c = + x – a c b = a + c Similarly, PF = a – a c x In hyperbola c > a ; and since P is to the right of the line x = a , x > a , c a x > a . Therefore, a – c a x becomes negative. Thus, PF = c a x – a.

CONIC SECTIONS Therefore PF – PF = a + c a x – cx a + a = a Also, note that if P is to the left of the line x = – a , then PF c – a  , PF = a – c x In that case P F – PF = a . So, any point that satisfies – b = , lies on the hyperbola. Thus, we proved that the equation of hyperbola with origin ( , ) and transverse axis along x -axis is – b = . A Note A hyperbola in which a = b is called an equilateral hyperbola .

Discussion From the equation of the hyperbola we have obtained, it follows that, we have for every point ( x , y ) on the hyperbola, b = + ≥ . i.e, a x ≥ , i.e., x ≤ – a or x ≥ a . Therefore, no portion of the curve lies between the lines x = + a and x = – a, (i.e. no real intercept on the conjugate axis).

Similarly, we can derive the equation of the hyperbola in Fig . (b) as b = These two equations are known as the standard equations of hyperbolas .