Mathematics

Session Parabola Session 2

Session Objective 1.Position of a point with respect to a parabola 2.Parametric form of parabola 3.Focal chord 4.Intersection of a line and a parabola 5.Tangent in various forms 6.Normal in various forms 7.Other Standard Parabolas

Position of a Point With Respect to a Parabola The point (h, k) lies outside, on, or inside the parabola y 2 = 4ax according as k 2 – 4ah >, =, < 0 S 1 is obtained by substituting point in the equation of curve S = 0, (S 1 is known as power of the point) Denote S  y 2 – 4ax and S 1  k 2 – 4ah then point is out, on or in as S 1 >, =, < 0

Parametric Form of Parabola y 2 = 4ax x = at 2, y = 2at where t is parameter General point on y 2 = 4 ax can be written as P(at 2, 2at) called ‘t’ point Parametric form of (y – k) 2 = 4a(x – h) is obtained as x – h = at 2, y – k = 2at i.e. x = h + at 2, y = k + 2at

Equation of Chord Joining Any Two Points on the Parabola Chord joining ‘t 1 ’, ‘t 2 ’ lying on y 2 = 4ax is given by Chord joining (x 1,y 1 ), (x 2,y 2 ) lying on y 2 = 4ax is given by Parametric Form Point Form

Condition of Focal chord and Focal distance Condition for focal chord (chord passing through focus) Chord joining 2 points Focal Distance of (x 1,y 1 ) on y 2 = 4 ax (distance from focus) Why? As above lines passes through (a,0) or

Length of Focal Chord Length of focal chord joining ‘t 1 ’, ‘t 2 ’ on y 2 = 4ax AB = length of focal chord = SA + SB

Intersection of a Line and a Parabola Let the line be y = mx + c and parabola be y 2 = 4ax then at point of intersection Real & distinct or Line cut parabola in 2 pts : c < a/m Real & Equal or Line Touches parabola: c = a/m Imaginary or Line can not intersect parabola: c > a/m Nature of Intersection

Equation of Tangent in Parametric(‘t’) Form Parametric Form x = at 2, y = 2at Equation of Tangent at ‘t 1 ’ i.e. (at 1 2,2at 1 ) Chord joining ‘t 1 ’, ‘t 2 ’ is (t 1 + t 2 )y = 2x + 2at 1 t 2 let t 2  t 1 where point of contact is (at 1 2,2at 1 ) and slope = 1/t 1 Alternative using Differentiation:

Equation of Tangent in Point(x 1,y 1 ) Form Point Form: Tangent at (x 1,y 1 ) on y 2 = 4ax Chord joining 2 points is (y 1 + y 2 )y = 4ax + y 1 y 2 let y 2  y 1 where point of contact is (x 1,y 1 ) and slope = 2a/y 1 Working Rule for Finding T = 0 T = 0 Replace x 2  xx 1, y 2  yy 1, x  (x+x 1 )/2 y  (y+y 1 )/2, xy  (xy 1 +x 1 y)/2 in the equation of curve

Equation of Tangent in Slope(m) Form Slope Form: Tangent of slope m to y 2 = 4ax y = mx + c intersect y 2 = 4ax in two real & equal points if c = a/m (done earlier) is the equation of tangent of slope m point of contact given by:

Point of Intersection of Two Tangents at y 2 = 4ax and Angle between them Parametric form Tangents at ‘t 1 ’, ‘t 2 ’ be Solving these we getx = at 1 t 2, y = a(t 1 + t 2 ) Angle between these tangents is given by

Equation of Normal in Parametric(‘t’) Form Parametric Form x = at 2, y = 2at Equation of Normal at ‘t 1 ’ i.e. (at 1 2,2at 1 ) Slope of tangent at ‘t 1 ’ is 1/t 1  slope of normal = –t 1 Alternative using Differentiation:  Equation is y – 2at 1 = –t 1 (x – at 1 2 ) with Foot of normal = ‘t 1 ’

Equation of Normal in Point(x 1,y 1 ) Form Point Form: Normal at (x 1,y 1 ) on y 2 = 4ax Slope of tangent = 2a/y 1  slope of normal = –y 1 /2a with foot of normal at (x 1,y 1 )

Equation of Normal in Slope(m) Form Slope Form: Normal of slope m to y 2 = 4ax Equation of Normal at ‘t 1 ’ i.e. (at 1 2,2at 1 ) As slope = m  –t 1 = m or t 1 = –m is the equation of normal in slope form with foot of normal at (am 2,–2am)

Point of Intersection of Two Normals at y 2 = 4ax Parametric form Normals at ‘t 1 ’, ‘t 2 ’ be Solving these we get

For Other Standard Parabolas Position of a Point y = 4ax 2 SOutInOnS 1 y = – 4ax 2 x = – 4ay 2 x = 4ay 2 S > 0 1 S = S > S < k – 4ah 2 k + 4ah 2 h – 4ak 2 h + 4ak 2

For Other Standard Parabolas Parametric Forms Parabolay 2 = 4axy 2 = –4axx 2 = 4ayx 2 = –4ay Co- Ordinates (at 2, 2at)(–at 2, 2at)(2at, at 2 )(2at, –at 2 ) Equations x = at 2 y = 2at x = –at 2 y = 2at x = 2at y = at 2 x = 2at y = –at 2

For Other Standard Parabolas Tangent in Point(x 1,y 1 ) Form Equation of parabola Tangent at (x 1, y 1 ) y 2 = 4axyy 1 = 2a(x + x 1 ) y 2 = –4axyy 1 = –2a(x + x 1 ) x 2 = 4ayxx 1 = 2a(y + y 1 ) x 2 = –4ayxx 1 = –2a(y + y 1 )

For Other Standard Parabolas Tangent in Parametric(‘t’) Form Equation of parabola Point of ContactTangent at ‘t’ y 2 = 4ax(at 2, 2at)ty = x + at 2 y 2 = –4ax(–at 2, 2at)ty = –x + at 2 x 2 = 4ay(2at, at 2 )tx = y + at 2 x 2 = –4ay(2at, –at 2 )tx = –y + at 2

For Other Standard Parabolas Tangent in Slope(m) Form Equation of parabola Point of contact Equation of tangent Condition of tangency y 2 = 4ax(a/m 2, 2a/m)y = mx+a/mc = a/m y 2 = –4ax(-a/m 2,-2a/m)y = mx–a/mc = –a/m x 2 = 4ay(2am, am 2 )y = mx–am 2 c = –am 2 x 2 = –4ay(–2am,–am 2 )y = mx+am 2 c = am 2

For Other Standard Parabolas Normals in Point(x 1,y 1 ) Form x 2 = –4ay x 2 = 4ay y 2 = –4ax y 2 = 4ax Normal at (x 1, y 1 ) Equation of parabola

For Other Standard Parabolas Normals in Parametric(‘t’) Form Equation of parabola Foot of NormalNormal at ‘t’ y 2 = 4ax(at 2, 2at)y+ tx = 2at+ at 3 y 2 = –4ax(–at 2, 2at)y– tx = 2at+ at 3 x 2 = 4ay(2at, at 2 )x+ ty = 2at+ at 3 x 2 = –4ay(2at, –at 2 )x– ty = 2at+ at 3

For Other Standard Parabolas Normals in Slope(m) Form Equation of parabola Foot of Normal Equation of Normal Condition of Normality y 2 = 4ax(am 2, –2am) y = mx–2am– am 3 c = –2am– am 3 y 2 = –4ax(–am 2, 2am) y = mx+2am+m 3 c = 2am+am 3 x 2 = 4ay (–2a/m, a/m 2 ) y = mx+2a+a/m 2 c = 2a+a/m 2 x 2 = –4ay (2a/m, –a/m 2 ) y = mx–2a–a/m 2 c = –2a–a/m 2

Other Method for Finding Tangents/Normals Let Curve be y = f(x) Tangent at (x 1,y 1 ) Normal at (x 1,y 1 )

Class Exercise

Class Exercise - 1 The line y = mx + 1 is a tangent to the parabola y 2 = 4x if (a) m = 1(b) m = 2 (c) m = 4(d) m = 3

Solution Condition of tangency of y = mx + c to y 2 = 4ax is Hence, answer is (a).

Class Exercise - 2 Show that x = my + c touches the parabola

Solution Substituting x, we get if line touches D = 0

Class Exercise - 3 Find the locus of the point of intersection of tangents at the extremities of a focal chord of y 2 = 4ax.

Solution Let be the extremities of the focal chord then t 1 t 2 = –2. Point of intersection of tangents is given by x = –a is the equation of directrix of the parabola Required locus is x = –a

Class Exercise - 4 If x + y = k is normal to y 2 = 12x then find k and the foot of the normal.

Solution y = –x + k, equation of normal in slope form is y = mx – 2am – am 3 with foot of normal as = = 9 and foot of normal is (3, 6) Note: Since, we know slope of normal, we have used slope form.

Class Exercise - 5 Prove that in a parabola semi-latus rectum is the harmonic mean of the segments of a focal chord.

Solution Let the equation of parabola be y 2 = 4ax Semi latus rectum = 2a Let chord joining ‘t 1 ’, ‘t 2 ’ be the focal chord points are Segments of the focal chord are

Solution contd.. H.M. of these two is = semi latus rectum Hence proved.

Class Exercise - 6 Consider a circle with its centre lying on the focus of the parabola y 2 = 2px such that it touches the directrix of the parabola. Then a point of intersection of the circle and the parabola is/are (a) (b) (c) (d)

Solution Equation of circle is

Solution contd.. For point of intersection with y 2 = 2px If is imaginaryReal points of intersections are Hence answer is (a), (b).

Class Exercise - 7 A is the point on the parabola y 2 = 4ax. The normal at A cuts the parabola again at the point B. If AB subtends a right angle at the vertex of the parabola, find the slope of AB.

Solution Let A be slope of normal at A is –t 1. Let normal again cuts the parabola y 2 = 4ax at

Solution contd.. From (i) and (ii) Slope of AB is

Class Exercise - 8 Find the locus of the point of intersection of those normals to the parabola y 2 = 8x which are at right angles to each other.

Solution Let the normals be drawn at the points their slopes are As normals are perpendicular t 1.t 2 =–1 Normals are given by Solving we get

Solution contd.. Solving we get as [from (ii)] [from (i)]

Class Exercise - 9 The locus of the mid point of the line segment joining the focus to a moving point on the parabola y 2 = 4ax is another parabola with directrix (a)x = –a(b) (c) x = 0(d)

Solution Focus (a, 0) Moving point be (h, k) such that k 2 = 4ah To find locus of mid point Hence, answer is (c).

Class Exercise - 10 Show that the locus of a point that divides a chord of slope 2 of the parabola y 2 = 4ax internally in the ratio 1 : 2 is a parabola. Find the vertex of this parabola.

Solution Let chord be the line joining the points Point that divides AB in 1 : 2 is

Solution contd.. From (i), (iii) we get

Solution contd.. Note:There could also be a point that divide AB internally in the ratio 2 : 1.