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Mathematics.

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Presentation on theme: "Mathematics."— Presentation transcript:

1 Mathematics

2 Ellipse Session - 2

3 Session Objectives

4 Session Objectives Equation of the tangent in point (x1, y1) form
Equation of normal in point (x1, y1) form Equation of tangent and normal in parametric form Number of tangents drawn from a point to an ellipse Director circle Equation of pair of tangents Equation of chord of contact Equation of normal in slope (m) form Number of normals (co-normal points) Equation of chord whose middle point is given Diameter of ellipse Conjugate diameters

5 Equation of the Tangent in Point (x1, y1) Form
Equation of tangent to ellipse at (x1, y1) is given by

6 Equation of the Tangent in Point (x1, y1) Form
Working rule for finding T = 0, replace and keeping constant unchanged.

7 Equation of Normal in Point (x1, y1) Form
Equation of normal at to ellipse given by

8 Equation of Tangent and Normal in Parametric () Form
Equation of tangent at using point form is Slope point of contact with the ellipse.

9 Equation of Tangent and Normal in Parametric () Form
Equation of normal at becomes Slope = foot of normal

10 Number of Tangents Drawn From a Point to an Ellipse
Two tangents can be drawn from a point to an ellipse it may be real or imaginary.

11 Number of Tangents Drawn From a Point to an Ellipse
Real and distinct if , i.e. point lies outside the ellipse. Real and coincident if , i.e. point lies on the ellipse. Imaginary if , i.e. point lies inside the ellipse.

12 Director Circle Director circle is the locus of point of intersection of perpendicular tangents to the conic. Equation of director circle of ellipse Let (h, k) be the point of intersection of tangents to and slope of tangent be m. Then we have

13 Director Circle if tangents are perpendicular, then , i.e.
Hence, locus of (h, k) is

14 Equation of Pair of Tangents
As we have seen earlier from a point (h, k) lying outside the ellipse, we have two real and distinct tangents possible. Combined equation of these tangents is given by or have usual meanings. Note: We can obtain above equation by eliminating ‘m’ from y – k = m(x – h) and

15 Equation of Chord of Contact
Equation of chord of contact of point (h, k) outside the ellipse is or T = 0 PQ is chord of contact of point (h, k).

16 Equation of Normal Slope (m) Form
Equation of normal of slope m to is Slope = m, foot of normal is Number of Normals (Co-normal Points) Four normals can be drawn from a point to an ellipse.

17 Equation of Chord Whose Middle Point is Given
Equation of chord of bisected at (h, k) is , i.e where T, S1 have usual meanings.

18 Diameter of Ellipse The locus of mid-point of a system of parallel chords of an ellipse is called diameter and chords are called its double ordinates. The end points of the diameter lying on the ellipse are called vertices of diameter. Equation of diameter of ellipse Let the system of parallel chords be given by y = mx + c, where ‘m’ is fixed and ‘c’ is a variable. Let (h, k) be its middle point. Then equation of chord with middle point at (h, k) is given by

19 Diameter of Ellipse Then its slope is m. Hence, Locus of (h, k) is or
which is the required equation of diameter. Note that diameter passes through the centre of ellipse. Hence, equation of diameter bisecting the parallel chords of slope ‘m’ of ellipse

20 Conjugate diameters Two diameters of an ellipse are said to be conjugate diameters. If each bisects the chords parallel to the other. Condition of conjugate diameters Let be two conjugate diameters of (Recall that diameter of ellipse passes through the centre of the ellipse.). Then diameter bisecting the chords parallel to is given by which is given as y= m2x.

21 Conjugate diameters Then or Thus, are conjugate diameters of ellipse .
Note: Major axis and minor axis are conjugate diameters, as each bisects the chords parallel to the other but product of their slopes is not defined.

22 Properties of conjugate diameters
The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by

23 Properties of conjugate diameters
(ii)The sum of the squares of any two conjugate semi-diameters (half of the diameter) is constant and is given by sum of squares of semi-axis, i.e. Note: That major axis and minor axis are also conjugate diameters. (iii)The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter passing through this point.

24 Properties of conjugate diameters
(iv) Tangents at the ends of the pair of conjugate diameters form a parallelogram, i.e. ABCD is a parallelogram.

25 Properties of conjugate diameters
(v) The area of the parallelogram formed by the tangents at the ends of conjugate diameters is constant and is given by the product of the axes, i.e. area (ABCD) = 4ab.

26 Concyclic Points Any circle intersects an ellipse in two or four real points. They are called concyclic points. If a, b, g, d be the eccentric angles of four concyclic points on an ellipse, then a + b + g + d = 2np, i.e. even multiple of p.

27 Class Exercise - 1 The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse is (a) sq. units (b) 9 sq. units (c) 27 sq. units (d) sq. units

28 Solution End points of latus rectum are given by

29 Solution contd.. Tangents at these points, using point form, are given by Equation of AD = 2x + 3y = 9 Area of the parallelogram ABCD = 4 × Area Hence, answer is (c).

30 Class Exercise - 2 Find the point on which is nearest to the line x + y = 7.

31 Solution The point which is nearest to the line is the point in the first quadrant, where tangent is parallel to PQ or if PQ is moved parallel to itself towards ellipse, the point where PQ first meets, i.e. touches the ellipse is the required point.

32 Solution contd.. Let be the required point, tangent at is Also lies on
(2, 1) is the required point. Note: (–2, –1) is the farthest from the line x + y = 7

33 Solution contd.. Alternative: Slope of tangent is –1.
Point of contact is Required point is (2, 1).

34 Class Exercise - 3 If the normal at the end of a latus rectum of an ellipse of eccentricity ‘e’ passes through one end of the minor axis, then (a) 1 (b) 2 (c) 3 (d) 4

35 Solution Slope of normal at Slope of normal at

36 Solution contd.. Equation of normal at is
If it passes through (0, –b), then be = ae – or ab = a2 – b2 = a2 e2

37 Solution contd.. Short cut: As e < 1 for ellipse
Now check with options. is the only possibility. Hence, answer is (a).

38 Class Exercise - 4 A tangent to the ellipse meets the ellipse in the points P and Q. Prove that the tangents at P and Q are at right angles.

39 Solution Let the tangents at P, Q intersect at R(h, k). Then according to the given conditions, chord of contact of R w.r.t touches i.e. is tangent to

40 Solution contd.. Hence, locus of (h, k) is
which is the equation of director circle of Hence, tangents at P, Q intersect at right angle.

41 Class Exercise - 5 Find the locus of mid-points of normal chords of the ellipse

42 Solution Let (h, k) be the mid-point. Then its equation is given by .
If it is normal at , then this equation is same as

43 Solution contd...

44 Class Exercise - 6 If P and D be the ends of semi-conjugate diameters, find the locus of foot of perpendicular from centre upon PD.

45 Solution Equation of PD is [Using equation of chord joining ]
Equation of line perpendicular to above line passing through (0, 0) is given by

46 Solution contd..

47 Class Exercise - 7 Find the locus of the middle points of chord of an ellipse which are drawn through the positive end of the minor axis.

48 Solution Let (a cos , b sin ) be the coordinate of the other extremities of the chord of ellipse The positive end of the minor axis is clearly (0, b). Let (x, y) be the mid-point of the chord.

49 Solution contd.. Squaring and adding (i) and (ii),
On simplifying the required locus of (x, y) is

50 Class Exercise - 8 Prove that the area of the triangle form by three points on an ellipse, whose eccentric angles are is Prove also that its area is to the area of the triangle formed by the corresponding points on the auxiliary circle as b : a.

51 Solution Let the coordinates of the given points on the ellipse be be
Area of the triangle formed by these points

52 Solution contd.. Using Second Part:
Clearly, coordinates of the points of the auxiliary circle corresponding to the given three points may be obtained by putting b = a.

53 Class Exercise - 9 Prove that the straight line is a normal to the ellipse if

54 Solution The normal at any point of the ellipse is given by
Comparing the given line, and the equation of the normal

55 Solution contd.. Squaring and adding

56 Class Exercise - 10 The tangents drawn from a point P to the ellipse make angles q1 and q2 with the major axis. Find the locus of P when q1 + q2 is constant = 2a.

57 Solution Let the equation of the tangent to the ellipse, and let P(h, k) lies on the tangent. Then

58 Solution contd..

59 Solution contd..

60 Thank you


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