Mathematics

Session Parabola Session 3

Session Objective 1.Number of Normals Drawn From a Point 2.Number of Tangents Drawn From a Point 3.Director circle 4.Equation of the Pair of Tangents 5.Equation of Chord of Contact 6.Equation of the Chord with middle point at (h, k) 7.Diameter of the Parabola 8.Parabola y = ax 2 + bx + c

Number of Normals Drawn From a Point (h,k) Parabola be y 2 = 4ax let the slope of the normal be m, then its equation is given by y = mx – 2am – am 3 if it passes through (h,k) then k = mh – 2am – am 3 i.e. am 3 + (2a – h)m + k = 0 This shows from (h,k) there are three normals possible (real/imaginary) as we get cubic in m

Observations from am 3 + (2a – h)m + k = 0 1.At least one of the normal is real as cubic equation have atleast one real root 2.The three feet of normals are called Co-Normal points given by (am 1 2, –2am 1 ), (am 2 2, –2am 2 ) and (am 3 2, –2am 3 ) where m i ’s are the roots of the given cubic eq n 3.Sum of the ordinates of the co-normal points = –2a (m 1 + m 2 + m 3 ) = 0 4.Sum of slopes of normals at co-normal points = 0 5.Centroid of triangle formed by co-normal points lies on axis of the parabola.

Observations from am 3 + (2a – h)m + k = 0 7. Thus we have following different cases arises: 3 real and distinct roots m 1, m 2, m 3 or m 1, m 2, –m 1 –m 2 3 real in which 2 are equal m 1, m 2, m 2 or –2m 2, m 2, m 2 3 real, all equal m 1, m 1, m 1 or 0, 0, 0  k = 0, h = 2a 1 real, 2 imaginary m 1,   i (  0)

Number of Tangents Drawn From a Point (h,k) Parabola be y 2 = 4ax let the slope of the tangent be m, then its equation is given by y = mx + a/m if it passes through (h,k) then k = mh + a/m i.e. hm 2 – km + a = 0 This shows from (h,k) there are two tangents possible (real/imaginary) as we get quadratic in m

Observations from hm 2 – km + a = 0 Discriminant = k 2 – 4ah = S 1 1.S 1 > 0 Point is outside parabola: 2 real & distinct tangents 2.S 1 = 0 Point is on the parabola: Coincident tangents 3.S 1 < 0 Point is inside parabola: No real tangent 4.m 1 + m 2 = k/h, m 1 m 2 = a/h

Director Circle Locus of the point of intersection of the perpendicular tangents is called Director Circle hm 2 – km + a = 0 m 1 m 2 = a/h = –1  h = –a i.e. locus is x = –a Hence in case of parabola perpendicular tangents intersect at its directrix. Director circle of a parabola is its directrix.

Equation of the Pair of Tangents Parabola be y 2 = 4ax then equation ofpair of tangents drawn from (h,k) is given by SS 1 = T 2 where S  y 2 – 4ax, S 1  k 2 – 4ah and T  ky – 2a(x + h) Pair of Tangents: (y 2 – 4ax)(k 2 – 4ah) = (ky – 2a(x + h)) 2

Equation of Chord of Contact Parabola be y 2 = 4ax then equation ofchord of contact of tangents drawn from (h,k) is given by T = 0 where T  ky – 2a(x + h) Chord of Contact is: ky = 2a(x + h)

Equation of the Chord with middle point at (h, k) Parabola be y 2 = 4ax then equation ofchord whose middle point is at (h,k) is given by T = S 1 where T  ky – 2a(x + h) and S 1  k 2 – 4ah Chord with middle point at (h,k) is: ky – 2a(x + h) = k 2 – 4ah i.e. ky – 2ax = k 2 – 2ah

Diameter of the Parabola Diameter: Locus of mid point of a system of parallel chords of a conic is known as diameter Let (h, k) be the mid point of a chord of slop e m then its equation is given by ky – 2ax = k 2 – 2ah if its slope is m then Locus is

Class Exercise

Class Exercise - 1 Find the locus of the points from which two of the three normals coincides.

Solution Let (h, k) be the point and m be the slope of the normal then As two normal coincides Let m 1 =m 2 then and

Solution contd..

Class Exercise - 2 Three normals with slopes m 1, m 2 and m 3 are drawn from a point P not on the axis of the parabola y 2 = 4x. If results in the locus of P being a part of the parabola, find the value of

Solution Let P be (h, k) then m i ’s are given by m 3 satisfies (i)

Solution contd.. locus of (h, k) is If it is a part of y 2 = 4x

Class Exercise - 3 Find the locus of the middle points of the normal chords of the parabola y 2 = 4ax.

Solution Let (h, k) be the middle point then its equation is given by T = S 1 i.e. ky – 2a (x + h) = k 2 – 4ah If it is also the normal of y 2 = 4ax then compare it with

Solution contd..

Class Exercise - 4 Find the locus of the point of intersection of the tangents at the extremities of chord of y 2 = 4ax which subtends right angle at its vertex.

Solution Let (h, k) be the point of intersection of tangents then its chord of contact is given by T = 0 i.e. ky – 2a (x + h) = 0...(i) Pair of lines are given by homogenising y 2 = 4ax using (i) Now according to the question pair of lines joining origin to the point of intersection of (i) with the parabola are at right angles.

Solution contd.. 4ax 2 – 2kxy + hy 2 = 0 Pair of lines are perpendicular if 4a + h = 0 hence locus of (h, k) is x + 4a = 0 Alternative: Let be the extremities of the chord. As chord subtends right angle at the vertex we have

Solution contd.. Point of intersection of tangents at these point is point becomes (–4a, a (t 1 + t 2 )) and its locus is x + 4a = 0

Class Exercise - 5 Find the equation of the diameter of the parabola given by 3y 2 = 7x, whose system of parallel chords are y = 2x + c.

Solution Let (h, k) be the middle point of the chord then its equation is given by T = S 1 Its slope is 2

Solution contd.. Alternative:

Class Exercise - 6 Three normals to the parabola y 2 = x are drawn through a point (c, 0), then (a) (b) (c) (d) None of these

Solution above equation have 3 real roots if 2 – 4c < 0 i.e. Hence,answer is (c).

Class Exercise - 7 The mid point of segment intercepted by the parabola x 2 =6y from the line x – y = 1 is ___.

Solution Let the mid point be (h, k), therefore, its equation is given by hx – 3 (y + k) = h 2 – 6k or hx – 3y = h 2 – 3k h = 3, k = 2 Hence (3, 2) is the mid point of the segment.

Class Exercise - 8 Draw y= –2x 2 + 3x + 1.

Solution

Class Exercise - 9 Find the locus of the point of intersection of tangents to y 2 = 4ax which includes an angle between them.

Solution Let (h, k) be the point of intersection then equation of pair of tangents are given by SS 1 = T 2

Solution contd..

Class Exercise - 10 Find the coordinates of feet of the normals drawn from (14, 7) to the parabola y 2 = 16x + 8y.

Solution Let the foot of the normal be then tangent at is given by

Solution contd.. which also passes through (14, 7) Corresponding Feet of normals are (0, 0), (8, 16) and (3, –4)