Exercise 10.2

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.

Compute the magnitude of the following vectors: = ˆ ˆ ;i j k+ + = ˆˆ ˆ2 7 3 ;i j k− − = 1 1 1 ˆˆ ˆ 3 3 3 i j k+ −

From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is

(A) $7$ cm    (B) $12$ cm    (C) $15$ cm    (D) $24.5$ cm

Find a vector in the direction of vector ˆˆ ˆ5 2i j k− + which has magnitude 8 units.

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Show that the vectors ˆ ˆˆ ˆ ˆ ˆ2 3 4 and 4 6 8i j k i j k− + − + − are collinear.

Prove that the parallelogram circumscribing a circle is a rhombus.

Find the direction cosines of the vector ˆˆ ˆ 2 3i j k+ + .

A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.

Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Show that the vector ˆˆ ˆi j k+ + is equally inclined to the axes OX, OY and OZ.

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ˆ ˆˆ ˆ ˆ ˆ2 and -i j k i j k+ − + + respectively, in the ratio 2 : 1 (i) internally (ii) externally

Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).

Show that the points A, B and C with position vectors, = ˆˆ ˆ3 4 4 ,i j k− − = ˆˆ ˆ2i j k− + and = ˆˆ ˆ 3 5i j k− − , respectively form the vertices of a right angled triangle.

In triangle ABC (Fig 10.18), which of the following is not true: (A) (B) (C) (D)

If are two collinear vectors, then which of the following are incorrect: (A) (B) (C) the respective components of are not proportional (D) both the vectors have same direction, but different magnitudes.

Write two different vectors having same magnitude.

In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to

(A) $60^\circ$    (B) $70^\circ$    (C) $80^\circ$    (D) $90^\circ$

Write two different vectors having same direction.

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80^\circ$, then $\angle POA$ is equal to

(A) $50^\circ$    (B) $60^\circ$    (C) $70^\circ$    (D) $80^\circ$

Find the values of x and y so that the vectors ˆ ˆ ˆ ˆ2 3 and i j xi yj+ + are equal.

x2 = - 16 y 5. y2 = 10x 6. x2 = - 9 y In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b (Fig10.22). 7. Focus (6,0); directrix x = - 6 8. Focus (0,-3); directrix y = 3 9. Vertex (0,0); focus (3,0) 10. Vertex (0,0); focus (-2,0) 11. Vertex (0,0) passing through (2,3) and axis is along x-axis. 12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis. 10. 5 Ellipse Definition 4 An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse (Fig10.20). ANote The constant which is the sum of the distances of a point on the ellipse from the two fixed points is always greater than the distance between the two fixed points. The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse(Fig 10.21). 10.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse (Fig 10.23). Take a point P at one end of the major axis. Sum of the distances of the point P to the foci is F1 P + F2P = F1O + OP + F 2P (Since, F1P = F 1O + OP) = c + a + a - c = 2a Take a point Q at one end of the minor axis. Sum of the distances from the point Q to the foci is F1Q + F 2Q = 2 22 2 c b c b+ + + = 2 22 c b+ Since both P and Q lies on the ellipse. By the definition of ellipse, we have 2 2 2c b+ = 2a, i.e., a = 2 2c b+ or a 2 = b2 + c2 , i.e., c = 2 2b a− . 10.5.2 Eccentricity Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., ce a= . Then since the focus is at a distance of c from the centre, in terms of the eccentricity the focus is at a distance of ae from the centre. 10.5.3 Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two such possible orientations are shown in Fig 10.24. We will derive the equation for the ellipse shown above in Fig 10.24 (a) with foci on the x-axis. (a) Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O be the origin and the line from O through F 2 be the positive x-axis and that through F1as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (- c, 0) and F2 be (c, 0) (Fig 10.25). Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a so given PF1 + PF 2 = 2a. ... (1) Using the distance formula, we have 2 22 2 ) () ( y c x y c x+ − + + + = 2a i.e., 2 2) ( y c x+ + = 2a - 2 2) ( y c x+ − Squaring both sides, we get (x + c)2 + y2 = 4a2 - 4 a 2 22 2 ) ( ) (y c x y c x+ − + + − 2 2 2 2 1x y a b + = which on simplification gives x a ca y c x− = + −2 2) ( Squaring again and simplifying, we get 2 2 2/2 c a y a x −+ = 1 i.e., 2 2/2 b y a x + = 1 (Since c 2 = a2 - b2) Hence any point on the ellipse satisfies 2/2 2/2 b y a x + = 1. ... (2) Conversely, let P (x, y) satisfy the equation (2) with 0 < c < a. Then y2 = b2       − 2 1 a x Therefore, PF1 = 2 2( )x c y+ + =       −+ + 2 2 2 2 2) ( a

= / 2 2 / 2 2 2 2( ) ( ) a xx c a c a   −+ + −    (since b2 = a2 - c2) =

a a   + = +   Similarly PF2 = ca x a− Hence PF1 + PF 2 = 2c ca x a - x aa a+ + = ... (3) So, any point that satisfies 2 2/2 b y a x + = 1, satisfies the geometric condition and so P(x, y) lies on the ellipse. Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is 2 2 2 2 x y a b + = 1. Discussion From the equation of the ellipse obtained above, it follows that for every point P (x, y) on the ellipse, we have 2/2 2/2 b y a x − = ≤ 1, i.e., x2 ≤ a2, so - a ≤ x ≤ a. Therefore, the ellipse lies between the lines x = - a and x = a and touches these lines. Similarly, the ellipse lies between the lines y = - b and y = b and touches these lines. Similarly, we can derive the equation of the ellipse in Fig 10.24 (b) as 2 2 2 2 1x y b a + = . These two equations are known as standard equations of the ellipses. ANote 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 (Fig10.24), we have the following observations:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7).

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Find the sum of the vectors = ˆˆ ˆ 2 ,i j k− + = ˆˆ ˆ2 4 5i j k− + + and = ˆˆ ˆ 6 - 7c i j k= − .

The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.

Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.

Find the unit vector in the direction of the vector = ˆˆ ˆ 2a i j k= + + .

A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$.

Find the unit vector in the direction of vector , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

For given vectors, = ˆˆ ˆ2 2i j k− + and = ˆˆ ˆi j k− + − , find the unit vector in the direction of the vector .

In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.