ncert books for class maths cbsc

A Note The standard equations of hyperbolas have transverse and conjugate axes as the coordinate axes and the centre at the origin. However, there are hyperbolas with any two perpendicular lines as transverse and conjugate axes, but the study of such cases will be dealt in higher classes. From the standard equations of hyperbolas (Fig11. ), we have the following observations: .

Hyperbola is symmetric with respect to both the axes, since if ( x , y ) is a point on the hyperbola, then (– x , y ), ( x , – y ) and (– x , – y ) are also points on the hyperbola. MATHEMATICS . The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.

For example, – has transverse axis along x -axis of length , while – has transverse axis along y-axis of length . . . Latus rectum Definition Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

As in ellipse, it is easy to show that the length of the latus rectum in hyperbola is b a . Example Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas: – = , (ii) y – x = Solution (i) Comparing the equation – = with the standard equation – b Here, a = , b = and c = b Therefore, the coordinates of the foci are ( ± , ) and that of vertices are ( ± , ).Also, The eccentricity e = c a = . The latus rectum b (ii) Dividing the equation by on both sides, we have – Comparing the equation with the standard equation – b = , we find that a = , b = and c b CONIC SECTIONS Therefore, the coordinates of the foci are ( , ± ) and that of the vertices are ( , ± ). Also, The eccentricity c e .

The latus rectum b Example Find the equation of the hyperbola with foci ( , ± ) and vertices ( , ± ). Solution Since the foci is on y-axis, the equation of the hyperbola is of the form – b Since vertices are ( , ± ), a = Also, since foci are ( , ± ); c = and b = c – a = . Therefore, the equation of the hyperbola is – = , i.e., y – x = . Example Find the equation of the hyperbola where foci are ( , ± ) and the length of the latus rectum is .

Solution Since foci are ( , ± ), it follows that c = . Length of the latus rectum = b or b = a Therefore c = a + b ; gives = a + a i.e., a + a – = , So a = – , . Since a cannot be negative, we take a = and so b = . Therefore, the equation of the required hyperbola is – = , i.e., y – x = MATHEMATICS Fig .

EXERCISE . In each of the Exercises to , find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. . – .

y – x = . In each of the Exercises to , find the equations of the hyperbola satisfying the given conditions. . Vertices ( ± , ), foci ( ± , ) .

Vertices ( , ± ), foci ( , ± ) . Vertices ( , ± ), foci ( , ± ) . Foci ( ± , ), the transverse axis is of length . .

Foci ( , ± ), the conjugate axis is of length . . Foci ( ± , ), the latus rectum is of length . .

Foci ( ± , ), the latus rectum is of length . vertices ( ± , ), e = . . Foci ( , ± ), passing through ( , ) Miscellaneous Examples Example The focus of a parabolic mirror as shown in Fig .

is at a distance of cm from its vertex. If the mirror is cm deep, find the distance AB (Fig . ). Solution Since the distance from the focus to the vertex is cm.

We have, a = . If the origin is taken at the vertex and the axis of the mirror lies along the positive x -axis, the equation of the parabolic section is y = ( ) x = x Note that x = . Thus y = Therefore y = ± Hence AB = y = × = cm. Example A beam is supported at its ends by supports which are metres apart.

Since the load is concentrated at its centre, there CONIC SECTIONS is a deflection of cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection cm? Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate axis be chosen as shown in Fig .

. Fig . The equation of the parabola takes the form x = ay . Since it passes through ,  , we have ( ) = a  , i.e., a = × = m Let AB be the deflection of the beam which is m.

Coordinates of B are ( x , ). Therefore x = × × = i.e. x = = metres Example A rod AB of length cm rests in between two coordinate axes in such a way that the end point A lies on x -axis and end point B lies on y -axis. A point P( x , y ) is taken on the rod in such a way that AP = cm.

Show that the locus of P is an ellipse. Solution Let AB be the rod making an angle θ with OX as shown in Fig . and P ( x , y ) the point on it such that AP = cm. Since AB = cm, we have PB = cm.

From P draw PQ and PR perpendiculars on y -axis and x -axis, respectively. Fig . MATHEMATICS From