Question 1

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. 2. 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 x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator. 10.5.4 Latus rectum Definition 6 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 10.26). To find the length of the latus r ectum of the ellipse 1x y a b 2 2 2 2+ = Let the length of AF2 be l. Then the coordinates of A are (c, l ),i.e., (ae, l ) Since A lies on the ellipse 2 2 2 2 1x y a b + = , we have 2 2 2 2 ( ) 1ae l a b + = ⇒ l 2 = b2 (1 - e 2) But 2 2 2 2 2 2 21c a - b be -

= = = Therefore l2 = 4/2 b a , i.e., 2bl a= Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is 22b a . Example 9 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 2 2 125 9 x y+ = Solution Since denominator of 2/25 x is larger than the denominator of 2/9 y , the major Fig 10. 26 axis is along the x-axis. Comparing the given equation with 2 2 2 2 1x y a b + = , we get a = 5 and b = 3. Also 2 2 25 9 4c a - b -= = = Therefore, the coordinates of the foci are (- 4,0) and (4,0), vertices are (- 5, 0) and (5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the eccentricity is 4 5 and latus rectum is 22 18 b a = . Example 10 Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36. Solution The given equation of the ellipse can be written in standard form as 2 2 14 9 x y+ = Since the denominator of 2/9 y is larger than the denominator of 2/4 x , the major axis is along the y-axis. Comparing the given equation with the standard equation 2 2 2 2 1x y b a + = , we have b = 2 and a = 3. Also c = 2 2a - b = 9 4 5- = and 5 ce a= = Hence the foci are (0, 5 ) and (0, - 5 ), vertices are (0,3) and (0, -3), length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the ellipse is 5 3 . Example 11 Find the equation of the ellipse whose vertices are (± 13, 0) and foci are (± 5, 0). Solution Since the vertices are on x-axis, the equation will be of the form 2 2 2 2 1x y a b + = , where a is the semi-major axis. Given that a = 13, c = ± 5. Therefore, from the relation c2 = a2 - b2, we get 25 = 169 - b2 , i.e., b = 12 Hence the equation of the ellipse is 2 2 1169 144 x y + = . Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ± 5). 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 2 2 2 2 1x y b a + = . Given that a = semi-major axis 20 102= = and the relation c 2 = a2 - b2 gives 52 = 102 - b 2 i.e., b2 = 75 Therefore, the equation of the ellipse is 2 2 175 100 x y+ = Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (- 1,4). Solution The standard form of the ellipse is 2 2/2 b y a x + = 1. Since the points (4, 3) and (-1, 4) lie on the ellipse, we have 19 16 2 2= +b a ... (1)

16 1 b a+ = 1 ….(2) Solving equations (1) and (2), we find that 2 247 7a = and 2 247 15b = . Hence the required equation is 2 2 1247247/157 x y + =      , i.e., 7x2 + 15y2 = 247.