Parabola Exercise 1 2,6 Q.1 (A) S(0, 1) directric x + 2y = 0 PS = PM. x y x y 2y 1 x 2y Q.2 (D) y 2 = 18 x. 2 = 3t. 2 t 3 Q.
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1 Prbol Exercise Q. (A) S(0, ) directric x + y = 0 PS = PM x y x y 5 5 x y y x y Q. (D) y = 8 x (t, t) t t = t t t,t, 4 9 4,6 Q. (C) y 4 x 5 Eqution of directrix is x + = 0 x 0 5 Q.4
2 y = 8x M P t,t x 0 S SP = PM = 8 + t = 8 t x coordinte = t () () = 6 Q.5 y = 4x ; s prbol psses through (-,) So 4 = 4 ( ) 4 4 L.R. 4 4 Q.6 (A) x 8y copre both of the Y 4AX 4A = 8; A = Y = x, X = y For focus X = A y = Y = 0 x = 0 S(0, ) Directrix is X + A = 0
3 ( y) + = 0 y = Q.7 (C) 6, Eqution of prbol x = 4by ; s prbol psses through (6,-) so 6 = 4b ( ) b = Eqution of prbol x = y Q.8 (A) x + 4x + y 7 = 0 x y..() Copre () with Y = 4AX..() Y = x +, X y, A Vertex X = 0 y 0 Y = 0 x + = 0 Vertex A, Q.9 4y 6x 4y = 5
4 5 y x y 4 5 y x 4 4 y x copre () with y = 4AX focus is X A x 8 Y 0 y 0 5 S, 8 Q.0 (C) y + 8y 8x = 0 (y + 4) = 8x + 6 (y + 4) = 8 (x + )..() Copre with Y = 4AX Eq n. of directrix X + A = 0 (x + ) + () = 0 x + 4 = 0 Q. (A) y 4kx + 8 = 0 y 4k x k vertex A,0 k
5 Perpendiculr distnce of vertex fro directrix x = 0 is equl to LR 4 k 4k 4 k k k k = k or k = k k + k = 0 or k k + = k or k =, Q. (C) A (, ), S (, ) AS =,, Eq n of prbol y 4x Q. (D) x = t, y = t eliinte t y x
6 y = 4 (x ) Q.4 y = 4x y 4x = 0 (S)(, ) = 9 4 > 0 (S)(,) = 4 < 0 P(, ) is outside the Prbol. Q (, ) is inside the prbol. Q.5 (D) P (,) : Q (,) R (, y) PQ QR y y R, R is interior point of y 4x = 0 So, , Q.6 (B) y x ; y x
7 4 L chord c For line which does not intersect c < Q.7 (C) y = 4x 4 = 4() = tngent t (, ) is yy = (x + x) y ( ) = (x + ) x + y + = 0 Q.8 (D) y =6x tngent in slope for 4 y x slope of tngent y x Q.9 (D) y = x ; slope of tngent = tn 45 o. =
8 point of contct, s, 4 Q.0 y = x + touches y = 6x 4, 4 Point of contct of tngent, 4 4, 4,4 Q. (A) Eqution of tngent t, t t x t y t put t u So, x = u, y = u which gives prbol y = 4x Eqution of tngent uy = x + u yx t t ty = t x + Q. (A) y = 4x
9 y x ; eq n of tngent in slope for & it psses through (-,-) then = 0 ( =, ) tn tn Q. (B) y = 4x..() x = 4by..() for point of intersection solve () & () x 4x 4b x = 0, x 64b x 4 b O P Point of intersection P4 b,4,b Angle between curves = ngle between the in tngents t intersection point slope y 4b
10 b x 4 b b b b tn b b b b o 60 Q.4 (C) (x + ) + y = 4 () Eq n of prbol y = 8x in pretric for is ty = x + t. () If () touches eq n () t 0 t t t t t 4 + t = t + t = 0, t =
11 t eq n of tngents re x = 0, y x 6 to be bove x xis slope of tngent is positive. Q.5 (B) Intersection point of y = 4x & x = 4y re A (0, 0), B (4, 4) eq n of line pssing through A & B is y = x () 6x + cy + d = 0.() () & () re se, so d = 0, b = c (b + c) = 0 Q.6 x + y = is norl. y = x + ; = y =x = = 9 Q.7 (B) = () ( ) () ( ), 4 Point t, t to y = 4x t
12 Eq n of norl is y + tx = t + t y x 8 8y + 4x = 8 + 8y + 4x = 9 Q.8 (B) y = 8x norl is prllel to y = x + 5 slope of norl foot of norl = (, ), 4, Q.9 (A) y = 4x end point of L.R. L (, ), L (, ) t =, t = eq n of norls re y + t x = t + t y + x = y x = cobined eqution of norl s re
13 [y + (x )] [y (x )] = 0 y (x ) = 0 Q.0 (A) x = 8y eq n of norl in pretric for x + ty = t + t x + ty = 4t + t () slope of norl (sy) t so put t in eq n () x y y x y = 4 eqution of norl in slope for if this norl psses through (h, k) so h + (4 k) + = 0 (,, ) () h But = (given) ( two of the norl s re perpendiculr) h The vlue of hs to stisfy eq n () 8 4 h 4 k 0 h h So locus (h, k) is x = (y 6) Q. y = 4x
14 t let (h, k) is centroid of P (t, t), Q (t, t), R (t t t, t) is k 0 so locus of centroid is y = 0 Q. (C) Two of the foot of norl s re (, ) & (, ) for y = 4x s we cn see tht foot of norl s given such tht t =, t = t + t + t = 0 + ( ) + t = 0 t = 0 so rd foot is (0, 0) Q. (B) (u, u), (v, v) re extreities of focl chord y = 4x So uv = (by property) Q.4 (C) y = 8x Eq n of chord of contct of (, 5) is T = 0 y5 x y x 5 5 Copre with y = x + c 4 8,c 5 5 Length of chord y = x + c for y = 8x is 4 c Q.5 (A) put the vlues now.
15 y = 4x P S (,0) Q (by property for focl chord) QS PS k b (given QS = k, PS = b) b k b Q.6 (B) y = x dieter of syste of prllel chords which re prllel to y = x + ( = ) is y 4 y Q.7 (C) y = 6x bsciss is = 4; ens the point is (4, 8) or (4, 8) P(4,8) T N(4,0) G L (sub tngent) = NT
16 eq n of tngent t (4, 8) is y (8) 8(x + 4) y = x + 8 T ( 4, 0) NT = 8 Q.8 (A) sub-tngent = sub-norl P(t, t) T ( t,0) N (t G( t,0),0) NT = NG (t ) = t P t,t, Q.9 (A) y = x () let, is pole of () w.r.t. y = x y x y x...() copre () & () 0
17 0, Q.40 (A) Let pole (h, k) so polr w.r.t circle x + y = r is xh + yk = r Eq n of tngent y = 4x is ty = x + t.().() () & () re se for se vlue of t so. h k r t t k r t, t h k k r h k locus of (h, k) is y r x Q.4 (A) Ry oving prllel to xis of the prbol then reflected ry hs to pss through focus of the prbol. y 4x focus is S (0, ) Q.4 (A) S (, b), directrix x y b Apply PS = PM x y b x y b b Q.4 (A)
18 y = x fter interchnging y to y, eq n of curve rein se so curve is syetric bout x ixs. Q.44 (B) y = 6x P (t, t) t = (t ) t = 0, Point is (4, 8) SP = PM = (4 + 4) = 8 Q.46 (B) 4x y 5 here so extreities of ltus rectu re Q.47 (C) vertex A (0, 0) directricx x + 5 = 0 so = 5 ltus rectu = 4 = 0 Q.48 (C) y = 6x vertex A (0, 0) the point, whose bsciss is 4, re 4,,, 5 5 eq n of line is y x
19 Q.49 (A) y = 4x P(t,t) R h,k Q t, t PR RQ t t t h t,k k h 4h = 9k So locus is 9y = 4x Q.50 (B) x y 5 x 4y 5 PS = PM where S is focus & M is perpendiculr on directrix. so locus of P is prbol where 5 (, 5) Q.5 (A) y = 8x S (, 0) if P(t, t) the SP = + t 4 = (+t ) t P, 4 Q.5 (C)
20 S (, 0) directric is x + 5 = 0 eq n of prbol is x 5 x y y = 4x Q.5 (A) y = 4x A O 0,0 B eq n of AB is lx + y + n = 0 eq n of pir of stright line OA & OB is given by hoogeniztion lx y So, y 4x 0 n Angle between lines is 90 o so 4l 0 n Q.55 (A) y = 4x eq n of tngent slope for is y x & this tngent psses through (0,) then () 0,
21 , 4 So, tn 4 ; put this vlue of in eq.() Q.56 (C) Point of intersection of tngents to y = 4x is (tt, (t + t)) ( () (), ( + )) = (, ) Q.57 (D) y = 4 (x ) eq n tngent is y x So 0 0 = + So ngle between tngents = Q.58 (B) y 4x = 0 (S)(0, ) = 4 0 > 0 & it psses through (0, 0) So point lies outside so rel & distinct tngents cn be drwn. Q.59 (B) y = 4x eq n tngent in slop for y x.() tngents re drwn fro, so
22 0.(), given tht = So, So, stisfy eq n () Q.60 (C) y + b = (x + ).() y + b = (x + ).() eq n () & () re tngents to the prbol y = 4x both tngents re drwn fro (, b) which lies on x =, so tngents drwn fro (-, -b) hs to be perpendiculr so Q.6 y = 4(x + ) eq n of tngent in slope for is
23 y x..() y = x + c () c Q.6 (B) y = 4x (, ) lies on directrix x = of the prbol. So, tngents drwn fro it re perpendiculr Q.6 (B) Intersection point of tngents t t, t t, which lies on xis y = 0 if (t + t) = 0 So, t = t Q.64 (D) y = 4x.() eq n of tngents ty = x + t x ty + t = 0 p p k 0 t t k 0 t t p p 4k Q.65 (A) kx + y = 4 y = x x..()..() () touches prbol () then
24 (4 kx) = x x x x (k + ) + 4 = 0 x x D 0 (k + ) 6 = 0 k =, 5 Q.66 (A) y = 4x x = 4by..()..() eq n of tngent for () ty = x + t..() x t x 4b t tx 4bx 4bt = 0 D = 0 6b + 4 (4b) t = 0 t b b t So eq n of tngent is b b x y 0 x b y b 0
25 Q.67 (A) x cos ysin p y cot x pcos ec..() y = 4(x + ) tngent in slope for is y x pcosec pcosec cot tn p cot 0 Q.68 (C) x + y =.() line () touches y y + x = 0 y y + ( y) = 0 (y ) = 0 y = So, x = 0 Point of contct is (0, ) Q.69 (B) x + y = y = 8x.().() eq n of tngent for () is ty = x + t x ty + t = 0..()
26 line () touches circle () then 0 0 t t (t ) = + t t 4 t = 0 t = t so eq n of tngents re y 0
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