PARABOLA EXERCISE 3(B)

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1 PARABOLA EXERCISE (B). Find eqution of the tngent nd norml to the prbol y = 6x t the positive end of the ltus rectum. Eqution of prbol y = 6x 4 = 6 = / Positive end of the Ltus rectum is(, ) =, Eqution of tngent yy = (x + x ) yy = (x + x ) y = x + y x = 0 is the eqution of tngent Slope of tngent is Slope of norml is Eqution of norml is x + y 9 = 0 y = x. Find the eqution of the tngent nd norml to the prbol x 4x 8y + = 0 t (4, /). Eqution of the the prbol is x 4x 8y + = 0. And point is (4, /) Eqution of tngents t (x, y ) is S =0 4x ( x + 4) 4 y + + = 0 x 4y = 0 x y = 0 Eqution of norml isy y = m(x x ) m-slope of norml Slope of tngent is / Slope of norml is. Therefore eqution of the norml is y = (x 4) y = 4x+ 6 4x + y 9 = 0

2 . Find the vlue of k if the line y = 5x + k is tngent to the prbol y = 6x. Eqution of the prbol is y = 6x Given line is y = 5x + k 5 k y= x+ Therefore m = 5, c = k 5 k y= x+ is tngent to y = 6x k / 6 c= = k = m 5/ 5 4. Find the eqution of the norml to the prbol y = 4x which is prllel to y x + 5 = 0. Given the prbol is y = 4x = Given line y x + 5 = 0 Slope m = The norml is prllel to the line y x+5 = 0 Slope of the norml = Eqution of the norml t t is y + tx = t + t slope = t = ( t = ) Eqution of the norml is x y = 0. y x = ( ) + ( ) = 4 8 = 5. Show tht the line x y + = 0 is tngent to the prbol y = 6x. Find the point of contct lso. Given prbol is y = 6x 4 = 6 = 4 Given line is x y + = 0 y = x + m =, c = 4 = = = c m Therefore given line is tngent to the prbol. Point of contct = 4 (4),, (,4) = = m m

3 6. Find the eqution of tngent to the prbol y = 6x inclined t n ngle 60 with its xis nd lso find the point of contct. II. Given prbol y = 6x Inclintion of the tngent is θ = 60 m = tn 60 = Therefore eqution of the tngent is y= x+ 4 y = x Point of contct =,, = m m y= mx+ m. Find the equtions of tngents to the prbol y = 6x which re prllel nd perpendiculr respectively to the line x y + 5 = 0. Find the coordintes of the points of contct lso. Given prbol is y = 6x 4 = 6 =4 Eqution of the tngent prllel to x y + 5 = 0 is y = x + c 4 Condition for tngency is c= = = m Eqution of the tngent is y = x + x y + = Point of contct is,, (,4) = = m m 4 Eqution of the tngent perpendiculr to x y + 5 = 0 is x+y+c =0 y = x c y = x c If bove line is tngent the c = /m 4 c= c= 6 Eqution of the perpendiculr tngent is y= x 8 y= x 6 x+ y+ 6= 0

4 Point of contct is m, m 4 8 =, = (6, 6). (/ 4) ( / ). If lx + my + n = 0 is norml to the prbol y = 4x, then show tht l + lm + nm = 0. Given prbol is y = 4x Eqution of the norml is y + tx = t + t tx + y (t + t ) = 0 Eqution of the given line is lx + my + n = 0 () () (), () re representing the sme line, therefore t (t+ t ) = = l m n t l = t = l m m (t+ t ) = m n n = t + t m l l l l + = + m m m m nm = l m + l l + lm + nm = 0. Show tht the eqution of common tngents to the circle x + y = nd the prbol y = 8x re y = ±(x + ). Given prbol y = 8x y = 4.x The eqution of tngent to prbol is mx my+ = 0 () y = mx + m. If () is tngent to the circle x + y =, then the length of perpendiculr from its centre (0, 0) to () is equl to the rdius of the circle. m 4 + m = 4 4= (m + m )

5 m 4 + m = 0 (m + )(m ) = 0 or m = ± Required tngents re y = ()x +, y = ( )x+ () ( ) y =± (x+ ) 4. Prove tht the tngents t the extremities of focl chord of prbol intersect t right ngles on the directrix. Let the prbol be y = 4x Eqution of the tngent t P(t ) is = + ty x t Eqution of the tngent t Q(t ) is = + t y x t Solving, point of intersection is T[tt,(t+ t )] Eqution of the chord PQ is (t+ t )y = x + tt Since PQ is focl chord, S (,0) is point on PQ. Therefore, 0 = + tt tt = -. Therefore point of intersection of the tngents is [,(t+ t )]. The x coordinte of this point is constnt. And tht is x = - which is the eqution of the directrix of the prbol. Hence tngents re intersecting on the directrix. 5. Find the condition for the line y = mx + c to be tngent to x = 4y. Eqution of the prbol is x = 4y.----() Eqution of the line is y = mx + c ----() Solving bove equtions, x = 4(mx + c ) x - 4mx -4 c =0 which is qudrtic in x. If the given line is tngent to the prbol, the roots of bove eqution re rel nd equl.

6 b -4c = 0 6 m +6c =0 m +c =0 c = - m is the required condition. 6. Three normls re drwn (k, 0) to the prbol y = 8x one of the norml is the xis nd the remining two normls re perpendiculr to ech other, then find the vlue of k. Eqution of prbol is y = 8x III. Eqution of the norml to the prbol is y + xt = t + t which is cubic eqution in t. therefore it hs roots. Sy t, t,t. where - t, -t,-t re the slopes of the normls. This norml is pssing through (k, 0) kt = t + t t + ( k)t = 0 t + ( k) = 0 Given one norml is xis i.e., x xis nd the remining two re perpendiculr. Thererfore m = 0=t, nd m m = ( t )( t ) =, t t = k = k = k = + = Eqution of the prbol is y = 8x 4 = 8 = k = = = 6.. If the norml t the point t on the prbol y = 4x meets it gin t point t then prove tht t t + t + = 0. Eqution of the prbol is y = 4x Eqution of norml t t = (t,t ) is y+xt = t + t. This norml meets the prbol gin t Therefore, ( ) ( ) t t = t t t ( ) = t t t tt + t + = 0 + = + t t t t t (t,t ).

7 . From n externl point P tngents re drwn to the prbol y = 4x nd these tngents mke ngles θ, θ with its xis such tht cotθ + cotθ is constnt d show tht P lies on horizontl line. Eqution of the prbol is y = 4x Eqution of ny tngent to the prbol is y= mx+ m This tngent psses through P(x, y ), then y = mx + m my = m x + m x my + = 0 let m, m be the roots of the eqution y m + m =, mm x = x where m nd m re the slopes of the tngents. m = tn θ nd m = tn θ Given cotθ + cotθ = d d tn θ + tn θ = m + m m m m m + = = m + m = d m m y x = d y = d x Locus of P(x, y ) is y = d which is horizontl line. d. Show tht the common tngent to the circle x + y = nd the prbol y = 4x intersect t the focus of the prbol y = 4x. Given prbol is y = 4x Let y= mx+ be the tngent. But this is lso the tngent to x + y = m Perpendiculr distnce from centre (0, 0) to the line = rdius 4 4 /m /m = = m + m + = (m + ) m = m + m m + m = 0

8 (m )(m + ) = 0 ( m + 0) m = 0 m=± Therefore, equtions of the tngents re y = -x nd y = x +. The point of intersection of these two tngents is ( -, 0) which is the focus of the prbol y = 4x. POLE AND POLAR THEOREM The eqution of the polr of the point P(x, y ) with respect to the prbol S = 0 is S = 0. Note. If P is n externl point of the prbol S = 0, then the polr of P meets the prbol in two points nd the polr becomes the chord of contct of P. Note. If P lies on the prbol S = 0, then the polr of P becomes the tngent t P to the prbol S = 0. Note. If P is n internl point of the prbol S = 0, then the polr of P does not meet the prbol. THEOREM The pole of the line lx + my + n = 0 (l 0) with respect to the prbol y n m = 4x is, l l. Proof :Eqution of the prbol is y = 4x Eqution of the line is lx + my + n = 0 () Let P(x, y ) be the pole The polr of P with respect to the prbol is S = 0 yy = (x + x ) x y y + x = 0 () Now () nd () represent the sme line. y x n m = = x =,y = l m n l l n m Pole P =, l l. Note. The pole of the line lx + my + n = 0 (m 0) with respect to the prbol x l n =4y is, m m.

9 CONJUGATE POINTS Note : The condition for the points P(x, y ), Q(x, y ) to be conjugte with respect to the prbol S = 0 is S = 0. CONJUGATE LINES Two lines L = 0, L = 0 re sid to be conjugte with respect to the prbol S = 0 if the pole of L = 0 lies on L = 0. THEOREM The condition for the lines l x + m y + n = 0 nd l x + m y + n = 0 to be conjugte with respect to the prbol y = 4x is l n + l n = m m. Proof : Eqution of the prbol is y = 4x Pole of the line l x + m y + n = 0 with respect to y n m = 4x is P,. l l Given lines re conjugte P lies on l x + m y + n = 0. n m l + m + n = 0 l l ln mm + ln = 0 ln + l n = mm MIDPOINT OF A CHORD THEOREM The eqution of the chord of the prbol S = 0 hving P(x, y ) s its midpoint is S = S. PAIR OF TANGENTS THEOREM The eqution to the pir of tngents to the prbol S = 0 from P(x, y ) is S = S S.

10 EXERCISE (C). Find the pole of the line x + y + 4 = 0 with respect to the prbol y = 8x. Eqution of the prbol is y = 8x 4 = 8 = Eqution of the given line is x + y + 4 = 0 l =, m =, n = 4 n m 4 Pole =, =, = (, 6) l l. Find the pole of x y 4 = 0 with respect to the prbol x 4x 8y + = 0. Given prbol is x 4x 8y + = 0. Let (, ) x y be the pole. Eqution of polr is S = 0 xx (x + x ) 4(y + y ) + = 0 x(x ) 4y x 4y+ = 0 (i) Compring eqution with eqution x y 4 = 0 x 4 x 4y+ We get : = = 4 x = 0, y = Pole P = (0, ).. Show tht the lines x y = 0 nd 6x y + = 0 re conjugte lines with respect to the prbol y = x. Eqution of the prábol is y = x Pole of the line is x y = 0 is Pole = 0, substitute 0, in 6x y + = = 0 0= 0 n m, 0 (/)( ) =, l l

11 Hence x y = 0 nd 6x y + = 0 re conjugte lines. 4. Find the vlue of k if x + y + 4 = 0 nd x + y + k = 0 re conjugte with respect to the prbol y = 8x. Eqution of the prbol is y = 8x = II. given x + y + 4 = 0 nd x + y + k = 0 re conjugte w.r.t. y = 8x. Therefore, l n + l n = m m ln + l n = k+ (4) m m = ()()() = k + 4 = k = 8 k = 4. Find the eqution of the chord of contct of the point A(, ) with respect to the prbol y = 4x. Find the points where the chord of contct meets the prbol. Using these find the equtions of tngents pssing through A to the given prbol. Given prbol is y = 4x = Eqution of chord of contct of A(, ) is S =0 y = (x + ) x y + 4 = (). x + 4 y = substitute this in y = 4x x + 4 = 4x x 5x + 4 = 0 (x 4)(x ) = 0 x = 4, y = 4, Eqution of the tngent t (4, 4) is S =0 y(4) = (x + 4) x y + 4 = 0 Eqution of the tngent t (, ) is S =0 y() = (x + ) x y + = 0.. Prove tht the polrs of ll points on the directrix of prbol y = 4x ( > 0) re concurrent t focus. Eqution of the prbol is y = 4x Eqution of the directrix is x = Any point on the directrix is P(, y ) Polr of P(, y ) is yy = (x ) This polr lwys psses through the fixed point (, 0) which is the focus of the prbol. The polrs of ll points on the directrix re concurrent t the focus of the prbol.

12 III.. If the polr of P with respect to the prbol y = 4x, touches the circle x + y = 4, then show tht P lies on the curve x y = 4. Eqution of the prbol is y = 4x Let P(x, y )be the pole. Polr of P(x, y ) is S =0 yy = (x + x ) x yy + x = 0 () If () is tngent to this circle x + y = 4 then Length of the perpendiculr form Centre is C(0, 0), = rdius of the circle x = 4 + y 0 0+ x = 4 + y Locus of P(x, y ) is x y = 4.. Show tht the poles of the chords of prbol y = 4x which subtend right ngle t vertex, lie on line prllel to its directrix. Eqution of the prbol is y = 4x Let P(x, y ) be the pole. Polr of P(x, y ) is S =0 yy = (x + x ) = x + x () yy x = x yy x = () x Homogenising () with the help of () y 4x(yy x) = 4x. = x x y = xyy 4x 4x y xy + x y = 0 but the lines re perpendiculr, therefore

13 Co efficient of x + Co efficient of y =0 4 + x = 0 x = 4 Locus of P(x, y ) is x = 4, which is prllel to the directrix is x =.. Show tht the chord of contct of ny point on the line x + 4 = 0 with respect to prbol y = 4x will subtends right ngle t the vertex. Eqution of the prbol is y = 4x -----() Any point on x + 4 = 0 is P( -4, y ) Eqution of the chord of contct of P is S =0 yy = x 8 8 = x yy yy = () 8 Homogenising () with help of () combined eqution of AQ, AR is 4x(x yy ) y = 4x.= y = x xyy 8 x xyy y = 0 From bove eqution, Coefficient of x + coefficient of y = = 0 AQR = 90 QR subtends right ngle t the vertex. 4. Show tht the poles of chords of the prbol y = 4x which re t constnt distnce from the focus lie on the curve y = 8x + 4x. Eqution of prbol is y = 4x Focus S = (,0) LET P(x, y ) be the pole. Polr of P(x, y ) is S =0 yy = (x + x ) = x + x x yy + x = 0

14 Given tht the perpendiculr distnce from S to this line = 0 + x + x = = 4 + y 4 + y 4 + y = 4( + x ) 4 + y = 4 + 4x + 8x Locus of P(x, y ) is y = 8x + 4x. PROBLEMS FOR PRACTICE. Find the coordintes of the vertex nd focus, nd the equtions of the directrix nd xes of the following prbols. i) y = 6x ii) x = 4y iii) x 9x + 5y = 0 iv) y x + 4y + 5 = 0. Find the eqution of the prbol whose vertex is (, ) nd focus is (, ). Ans. (x ) = (y + ). Find the coordintes of the points on the prbol y = x whose focl distnce is 5/. Ans. (, ) nd (, ) 4. Find the eqution of the prbol pssing through the points (, ), (, ) nd (, ) nd hving its xis prllel to the x-xis. Ans. 7y y + 6x 6 = 0 5. A double ordinte of the curve y = 4x is of length 8. Prove tht the line from the vertex to its ends re t right ngles. Let P = (t, t) nd P = (t, t) be the ends of double ordinte PP. Then 8 = PP = 0 + (4t) = 4t t = P = (4, 4), P = (4, 4) Slope of AP slope of AP 4 4 = = 4 4 π PAP =

15 6. (i) If the coordintes of the ends of focl chord of the prbol y = 4x re (x, y ) nd (x, y ), then prove tht x x =, y y = 4. (ii) For focl chord PQ of the prbol y = 4x, if SO = l nd SQ = l then prove tht + =. l l i) Let P(x,y ) = (t, t ) nd Q(x, y ) = (t, t ) be two end points of focl chord. P, S, Q re colliner. Slope of PS = Slope of QS t t = t t = t t t t t t tt (t t) + (t t) = 0 + t t = 0 t t = From () = = = xx tt (t t) = = = y y t t 4 (t t ) 4 ii) Let P(t, t ) nd Q(t, t ) be the extremities of focl chord of the prbol, then t t = (from()) l = SP = (t ) + (t 0) = (t ) + 4t = (+ t ) l = SQ = (t ) + (t 0) = (t ) + 4t = (+ t ) ( l )( l ) = t t = (t t ) = [ tt = ] ll ( l + l ) = 0 + = l l 7. If Q is the foot of the perpendiculr from point P on the prbol y = 8(x ) to its directrix. S is the focus of the prbol nd if SPQ is n equilterl tringle then find the length of side of the tringle. Ans. 8

16 8. Find the condition for the stright line lx + my + n = 0 to be tngent to the prbol y = 4x nd find the coordintes of the point of contct. n m Ans., l l 9. Show tht the stright line 7x + 6y = is tngent to the prbol y 7x 8y + 4 = 0 nd find the point of contct. Ans. (, ) 0. Prove tht the norml chord t the point other thn origin whose ordinte is equl to its bsciss subtends right ngle t the focus. Let the eqution of the prbol be y = 4x nd P(t, t) be ny point On the prbol for which the bsciss is equl to the ordinte. i.e. t = t t = 0 or t =. But t 0. Hence the point (4, 4) t which the norml is y + x = () + () y = ( x) Substituting the vlue of () y = ( x) in () we get ( x) = 4x x x + 6 = (x 4)(x 9) = 0 x = 4, 9 Corresponding vlues of y re 4 nd 6. () Hence the other points of intersection of tht norml t P(4, 4) to the given prbol is Q(9, 6), we hve S(, 0). Slope of the SP = m = = Slope of the SQ = m = = 9 4 Clerly m m =, so tht SP SQ.. From n externl point P, tngent re drwn to the prbol y = 4x nd these tngent mke ngles θ, θ with its xis, such tht tnθ + tnθ is constnt b. Then show tht P lies on the line y = bx.

17 . Show tht the common tngent to the prbol y = 4x nd x = 4by is / / / / x + yb + b = 0. The equtions of the prbols re y = 4x x = 4by () nd () Eqution of ny tngent to () is of the form y= mx+ () m If the line () is tngent to () lso, we must get only one point of intersection of ()nd (). Substituting the vlue of y from () in (), we get x = 4b mx+ m is mx 4bm x 4b=0 should hve equl roots therefore its discrimi-nent must be zero. Hence 6b m 4 4m( 4b) = 0 6b (bm 4 + m) = 0 m(bm + ) = 0, but m 0 m = / b / substituting in () the eqution of the common tngent becomes / y= x+ b b / or / / / / x+ b y+ b = 0.. Prove tht the re of the tringle formed by the tngents t (x, y ), (x, y ) nd (x, y ) to the prbol y = 4x ( > 0) is (y y )(y y )(y y ) sq.units. 6 Let D(x, y ) = (t,t ) E(x,y ) = (t,t ) nd = F(x, y ) (t, t ) Be three point on the prbol. y = 4x ( > 0) The eqution of the tngents t D, E nd F re = + t y x t...() = + t y x t...() = + t y x t...() () () (t t )y = (t t )(t + t ) y = (t + t ) substituting in () we get,

18 x = t t The point of intersection of the tngents t D nd E is sy P[t t, (t +t )] Similrly the points of intersection of tngent t E, F nd t F, D re Q[t t, (t +t )] nd R[t t, (t +t )] respectively. Are of PQR = Absolute vlue of = Absolute vlue of = Absolute vlue of = Absolute vlue of tt (t + t ) t t (t + t ) t t (t + t ) tt t + t tt t+ t tt t + t t(t t) t t 0 t(t t) t t 0 tt t + t t 0 (t t )(t t ) t 0 tt t + t = (t t )(t t )(t t ) = (t t )(t t )(t t ) 6 = (y y )(y y )(y y ) sq. units Find the vlue of k if i) Points (, ) (k ) re conjugte with respect to the prbol y = 8x. ii) The line x + y + = 0 nd x y + k = 0 re conjugte with respect to the prbol y + 4x y = 0. Ans. (i) /, (ii) 5. Prove tht the poles of norml chord of the prbol y = 4x lie on the curve(x + )y + 4 = 0.

19 6. Prove tht the poles of tngents to the prbol y = 4x with respect to the prbol y = 4bx lie on prbol. Eqution of ny tngent to y = 4x is of the form y= mx+ () m Let P(x, y ) be the pole of () w.r.t. y = 4bx Then the polr of P(x, y ) w.r.t y = 4bx is : yy = b(x + x ) () nd () represent the sme line Compring the coefficients y b bxm b = = m = m= m x y Eliminting m, = y = x x 4b 4b y The pole P(x, y ) lies on the prbol is : 4b y = x 7. If the norml t t nd t to the prbol y = 4x meet on the prbol, then show tht t t =. Proof : Let the normls t t nd t meet t t on the prbol. The eqution of the norml t t is : y + xt = t + t () Eqution of the chord joining t nd t is : y(t + t ) = x + t t () () nd () represent the sme line t+ t = t = t t t Similrly t = t t t = t t t = t t t t (t t ) t t = t t =. tt

1. If y 2 2x 2y + 5 = 0 is (A) a circle with centre (1, 1) (B) a parabola with vertex (1, 2) 9 (A) 0, (B) 4, (C) (4, 4) (D) a (C) c = am m.

1. If y 2 2x 2y + 5 = 0 is (A) a circle with centre (1, 1) (B) a parabola with vertex (1, 2) 9 (A) 0, (B) 4, (C) (4, 4) (D) a (C) c = am m. SET I. If y x y + 5 = 0 is (A) circle with centre (, ) (B) prbol with vertex (, ) (C) prbol with directrix x = 3. The focus of the prbol x 8x + y + 7 = 0 is (D) prbol with directrix x = 9 9 (A) 0, (B)