Parabola. The fixed point is called the focus and it is denoted by S. A (0, 0), S (a, 0) and P (x 1, y 1 ) PM=NZ=NA+AZ= x 1 + a

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: Conic: The locus of a point moving on a plane such that its distance from a fixed point and a fixed straight line in the plane are in a constant ratio é, is called a conic.

. Equation of a parabola in standard form Proof: Let S be the focus, l=0 be the directrix of the parabola. Let P (x, y ) be a point on the parabola.

1 : Conic: The locus of a point moving on a plane such that its distance from a fixed point and a fixed straight line in the plane are in a constant ratio é, is called a conic. The fixed point is called the focus and it is denoted by S. The fixed straight line l is called the directrix.. Equation of a parabola in standard form Proof: Let S be the focus, l=0 be the directrix of the parabola. Let P (x, y ) be a point on the parabola. Let M, Z be the projections of the P, S on l=0 respectively. Let N be the projection of p on SZ. Let A be the midpoint of SZ, SA=AZ Let SA=AZ=a, A is called vertex of the parabola. Take AS as X-axis and AY as Y-axis A (0, 0), S (a, 0) and P (x, y ) PM=NZ=NA+AZ=x + a The constant ratio e is called the eccentricity. The straight line of the plane passing through the focus and perpendicular to the directrix is called the axis. SP = e Where PM is the perpendicular distance PM from P the directrix. If e=, the conic is called a parabola. If 0<e<, the conic is called ellipse. If e>, the conic is called hyperbola. From the definition of the parabola SP PM = e= S.O.B SP = PM (x a) + (y 0) = (x + a) x + a ax + y = x + a + ax y = 4ax Thus the equation of the parabola in standard form is y = 4ax

2 . Find the coordinates of the vertex and focus, the equation of the directrix and axis of the parabola y x + 4y + 5 = 0. Sol: the given equation is y x + 4y + 5 = 0 y + 4y = x 5 y + 4y + = x 5 + (y + ) = (x ) [y ( )] = (x ) Comparing with[y k] = 4a(x h) We get, 4a= a=/4 and (h, k) = (, -) i. Vertex (, -) ii. Focus(h+a, k)=(+ 4, ) iii. =( 5, ) Equation of the directrix is x=h-a x = 4 = 3 4 4x 3 = 0 iv. Length of the latus rectum is 4a=. v. Eq n of axis y β = 0 y + = 0 3. Find the coordinates of the vertex and focus, the equation of the directrix and axis of the parabola x x + 4y 3 = 0. Sol: the given equation is x x + 4y 3 = 0 x x = 4y + 3 x x + = 4y (x ) = 4(y ) [x ] = 4(y ) 4. Find the equation of the parabola whose axis is parallel to the y-axis and passing through the points (4, 5), (-, ), (-4, ). Sol: The equation of the parabola whose axis is parallel to the y-axis y = ax + bx + c ( ) (4, 5) lies on (*) 5 = a(4) + b(4) + c 5 = 6a + 4b + c () (-, ) lies on (*) = a( ) + b( ) + c = 4a b + c () (-4, ) lies on (*) = a( 4) + b( 4) + c = 6a 4b + c (3) Solving () & () solving () & (3) 5 = 6a + 4b + c 5 = 6a + 4b + c = 4a b + c = 6a 4b + c 6 = a + 6b. (4) -6=8b (5) b = Sub b = in (4) 6 = a a = 6 + = 6 a = 6 = Sub a, b in () 5= c c = 5 Substituting the values of a,b and c in (*) we get the required equation of the parabola y = x + ( )x + 5 x y 4x + 0 = 0. Comparing with[x h] = 4a(y k) We get, 4a=4 a= and (h, k) = (, ). Vertex (, ). Focus(h, k-a)=(, )=(, 0) 3. Equation of the directrix is y=k+a y = + = y = 0 = 0 4. Length of the latus rectum is 4a=4. 5. Eq n of axis x α = 0 x = 0

3 5. Find the equation of the parabola whose axis is parallel to the x-axis and passing through the points (-, ), (, ), (-, 3). Sol: Sol: The equation of the parabola whose axis is parallel to the y-axis x = ay + by + c ( ) (-, ) lies on (*) = a() + b() + c = a + b + c () (, ) lies on (*) = a() + b() + c = 4a + b + c () (-, 3) lies on (*) = a(3) + b(3) + c = 9a + 3b + c (3) Solving () & () solving () & (3) = a + b + c = a + b + c = 4a + b + c = 9a + 3b + c 3 = 3a b. (4) -=-8a-b (5) Solving (4) & (5) (a, b)=[ +6, ] = [ 5, ] = [ 5, ] Substituting the values of a,b in () we get ( 5 ) + ( ) + c = 8 + c = c = 0 Substituting the values of a,b and c in (*) we get the required equation of the parabola y = 5 x + ( ) x 0 5x + y + x + 0 = Show that the equation of common tangents to the circle x + y = a and the parabola y = 8ax are y=±(x + a). Sol: Given equation of the circle x + y = a () [x + y = ( a), y = 8ax.. () [y = 4a x, a = a] r = a] Let m be the slope of common tangent. Equation of tangent to (), y = mx ± a + m (3) [ y = mx ± r + m is eq n of tangent to x + y = r ] Equation of tangent to (), y = mx + a..... (4) m a y = mx + [ m is eq n of ] tangent to y = 4a x (3), (4) Represents same line ± a + m = a m squaring on both sides a ( + m )= 4a m m ( + m )= m + m 4 = 0 m 4 + m m = 0 m (m + )-(m + )=0 (m )(m + )=0 m = 0, or m + = 0 m = or m = is not possible m = ± sub in (4) y = ±mx + a ± y=±(x + a). 3

4 3x y + = 0 7. Find the equation of the parabola whose focus is (-, 3) and directrix is the line x+3y-4=0. Also find the length of the latex rectum and the equation of the axis of the parabola. Sol: Given S (-, 3) Eq n of directrix l=x + 3y 4 = 0. Let P (x, y ) be a point on the parabola. Draw a perpendicular PM on the to the line L=0. SP = SP = PM { lar dist = ax +by +c } PM a +b (x + ) + (y 3) = x +3y 4 +3 S.O.B (x + ) + (y 3) = (x +3y 4) 3 3[x x + y + 9 6y ] =[4x + 9y x y 4y 6x ] 3x x + 3y y -4x 9y 6 x y + 4y + 6x = 0 9x x y +4y + 68x 54y + 53 = 0 Locus of P is 9x xy+4y + 68x 54y + 53 = 0 is the eq n of the parabola. 8. Prove that the area of the triangle inscribed in the parabola y = 4ax is 8a (y y )(y y 3 )(y 3 y ) Sq.uints wherey, y, y 3 are ordinates of its vertices. Sol: Given eq n of the parabola y = 4ax.() Let A (x, y ) = (at, at ); B (x, y ) = (at, at ) C (x 3, y 3 ) = (at 3, at 3 ) A lies on () y = 4ax x = y ABC In scribed in the parabola = x x y y x x 3 y y 3 y = ) y (y 4a y ) y y 3 ( y y 4a = 8a (y y )(y + y ) (y y ) (y y 3 )(y + y 3 ) (y y 3 ) = (y y )(y y ) (y + y ) 8a (y + y 3 ) = (y y )(y y ) y 8a + y y y 3 8a (y y )(y y 3 )(y 3 y ) Sq.uints 4a Length of latus rectum=4a=a = sz =[ lar dist from S(, 3) to l = 0] = ( )+3(3) 4 = = = Eq n of the axis is the line lar to the direcrtix and passing through S(, 3) is b(x x ) a(y y ) = 0 3(x + ) (y 3) = 0 3x + 6 y + 6 = 0 4

5 9. Prove that the area of the triangle formed by the tangents at (x, y ), (x, y ), (x 3, y 3 ) to the y = 4ax is 6a (y y )(y y 3 )(y 3 y ) Sq.uints wherey, y, y 3 are ordinates of its vertices. Sol: Given eq n of the parabola y = 4ax.() Let D (x, y ) = (at, at ); E (x, y ) = (at, at ) F (x 3, y 3 ) = (at 3, at 3 ) Point of intersections of tangents are A = [at t, a(t + t ) ]; B = [at t 3, a(t + t 3 ) ] C = [at 3 t, a(t 3 + t ) ] 0. If a normal chord at point t on the parabola y = 4ax subtends a right angle at vertex, then prove that t = ± Sol: Given eq n of the parabola y = 4ax. () Eq n of the normal at t on the parabola is y+xt y + xt = at + at 3 =. () at+at 3 homogenising () Using () we get, y = 4ax() y = 4ax( y+xt at+at 3) y = 4x( y+xt t+t 3) y (t + t 3 ) = 4x(y + xt) y (t + t 3 ) = 4xy + 4x t 4tx y (t + t 3 ) + 4xy = 0 ABC In scribed in the parabola = x x y y x x 3 y y 3 = at t at t 3 at + at at at 3 at t 3 at 3 t at + at at 3 at Subtends a right angle coeff x coeff y = 0 4t t t 3 = 0 t t 3 = 0 t( t ) = 0 t = 0 and t = t = ± = a t (t t 3 ) (t t 3 ) t 3 (t t ) (t t ) = a (t t 3 )(t t ) t t 3 = a (t t 3 )(t t ))t t 3 = a (t t )(t t 3 )(t 3 t ) Sq.uints y = at t = y a = a (y y ) a. (y y 3 ) a. (y 3 y ) Sq.uints a 6a (y y )(y y 3 )(y 3 y ) Sq.uints 5

Conic Sections. Geometry - Conics ~1~ NJCTL.org. Write the following equations in standard form.

Conic Sections. Geometry - Conics ~1~ NJCTL.org. Write the following equations in standard form. Conic Sections Midpoint and Distance Formula M is the midpoint of A and B. Use the given information to find the missing point. 1. A(, 2) and B(3, -), find M 2. A(5, 7) and B( -2, -), find M 3. A( 2,0)