# Hyperbola-02-Solved example

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Example 1 : SOLVED SUBJECTIVE EXAMPLES x 2  y2  x 2  y2  A tangent to the hyperbola a 2 locus of the midpoint of PQ. Solution : b2 1 cuts the ellipse a 2 b2 1 in points P and Q. Find the x 2  y2  Let M(x1, y1) be the midpoint of the chord PQ of the ellipse a 2 b2 1. xx yy x2 y2 b2xx b2 x2 y2  Equation of PQ is 1  1  1  1  y   1   1  1  a2 b2 a2 b2 a2y y a2 b2  This is tangent to the hyperbola x 2  y2  b4  x2 y2 2 b4 x2 a 2 b2 1  x2 y2 2 x 2 y2 if  1  1   a 2 1  b2   1  1   1  1 y2  a 2 b2  a 4 y2  x2 y2 2  a 2 x2 b2  y2 a 2 b2 Hence locus of (x1, y1) is        a 2 b2 Example 2 : A straight line is drawn parallel to the conjugate axis of the hyperbola x 2  y2  to meet it and a 2 b2 1 the conjugate hyperbola respectively in the point P and Q. Show that the normals at P and Q to the curves meet on the x-axis. Solution : Let P(a sec  , b tan ) be a point on the hyperbola, and Q(a tan , b sec  ) be a point on the conjugate hyperbola.  a sec  = atan  sec  = tan Equation of the normal to the hyperbola x 2  y2   a tan  a 2 b2 1 at P is y – b tan = bsec  (x – b sec  ) Equation of the normal to the conjugate hyperbola at Q is y – b sec  = – Eliminate x and use sec = tan We get y (sec  – tan ) = 0  y = 0 Hence the normals meet on the x-axis. a sec  (x a tan ) b tan Example 3 : From a point G on the transverse axis of the hyperbola x  y2 a 2 b2 1, GL is drawn perpendicular to one of its asymptotes. Also Gp is a normal to the curve at P. Prove that LP is parallel to the conjugate axis. Solution : Let P(a sec  , btan ) be any point on the hyperbola Equation of the normal at P is ax cos  + by cot  = a2 + b2. It meets the x-axis (transverse axis) at y = 0 2 2 a2  b2   x = a  b sec  G   a sec , 0 a   The equation of line perpendicular to the asymptote bx – ay = 0 and passing through G, i.e., equation of GL is y = – a x b   a 2  b2 a sec   ax + by = (a2 + b2) sec  Its intersection with the asymptote bx – ay = 0 gives x = a sec  . So the x coordinate of L is a sec  , which is equal to the x-coordinate of the point P  LP is parallel to the y-axis  LP is parallel to the conjugate axis. Example 4 : A variable straight line of slope 4 intersects the hyperbola xy = 1 at two points. Find the locus of the point which divides the line segment between these points in the ratio 1 : 2. [IIT-1997] Solution : Let the line be y = 4x + c. It meets the curve xy = 1 at x (4x + c) = 1  4x2 + cx –1  x + x2 = –c/4 Also y (y – c) = 4  y2 – cy – 4 = 0  y + y = c 2 Let the point which divides the line segment in the ratio 1 : 2 be (h, k)  Also x1  2x2  h 3 y1  2y2  k 3  x2  y2 = 3h + c/4  x1 = 3k – c  y1 = –c/2 – 3h = –3k + 2c Now (h, k) lies on the line y = 4x + c  k

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### Hyperbola-02-Solved example

• 1. SOLVED SUBJECTIVE EXAMPLES Example 1 : A tangent to the hyperbola 1 b y a x 2 2 2 2   cuts the ellipse 1 b y a x 2 2 2 2   in points P and Q. Find the locus ofthemidpoint ofPQ. Solution : Let M(x1 , y1 ) be the midpoint ofthe chord PQ ofthe ellipse 1 b y a x 2 2 2 2   . Equation ofPQ is 2 2 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 xx yy x y b xx x y b y a b a b a y y a b               This is tangent to the hyperbola 1 b y a x 2 2 2 2   if 2 2 1 4 2 1 4 2 2 2 2 1 2 2 1 2 1 4 b y a x b a b y a x y b             2 2 1 2 2 1 2 2 2 1 2 2 1 b y a x b y a x            Hence locus of(x1 , y1 ) is 2 2 2 2 2 2 2 2 2 b y a x b y a x            Example 2 : A straight line is drawn parallelto the conjugate axis ofthe hyperbola 1 b y a x 2 2 2 2   to meet it and the conjugate hyperbola respectivelyinthepoint P and Q. Show that the normalsat P and Qto the curves meet on the x-axis. Solution : Let P(a sec, b tan) be a point on the hyperbola, and Q(a tan , b sec ) be a point on the conjugate hyperbola.  a sec = atan  sec = tan Equationofthe normalto the hyperbola 1 b y a x 2 2 2 2   at P is y – b tan =    sec b tan a (x – b sec) Equation ofthe normal to the conjugate hyperbola at Q is y– b sec = – ) tan a x ( tan b sec a     Eliminate x and use sec = tan We get y (sec – tan ) = 0  y = 0 Hence the normals meet on the x-axis.
• 2. Example 3 : Froma point G onthe transverse axis ofthe hyperbola 1 b y a x 2 2 2 2   , GLis drawnperpendicular to one of its asymptotes. Also Gp is a normal to the curve at P. Prove that LP is parallel to the conjugate axis. Solution : Let P(a sec, btan) be anypoint on the hyperbola Equation of the normalat P is ax cos + bycot  = a2 + b2 . It meets the x-axis (transverse axis) at y= 0  x =   sec a b a 2 2  2 2 a b G sec ,0 a          The equation of line perpendicular to the asymptote bx – ay = 0 and passing through G, i.e., equation ofGL is y= –            sec a b a x b a 2 2  ax + by = (a2 + b2 ) sec Its intersection with the asymptote bx – ay = 0 gives x = a sec. So the x coordinate of L is a sec, which is equalto the x-coordinate of the point P  LP is parallelto the y-axis  LP is parallelto the conjugate axis. Example 4 : Avariablestraight line ofslope4 intersects the hyperbolaxy=1 at two points. Find thelocus ofthe point which divides the line segment betweenthese points in the ratio 1 : 2. [IIT-1997] Solution : Let the line be y = 4x + c. It meets the curve xy= 1 at x (4x + c) = 1  4x2 + cx –1  x1 + x2 = –c/4 Also y (y – c) = 4  y2 – cy – 4 = 0  y1 + y2 = c Let the point which divides the line segment in the ratio 1 : 2 be (h, k)  h 3 x 2 x 2 1    x2 = 3h + c/4  x1 = –c/2 – 3h Also k 3 y 2 y 2 1    y2 = 3k – c  y1 = –3k + 2c Now (h, k) lies on the line y = 4x + c  k = 4h + c  c = k – 4h  x1 = –k/2 + 2h – 3h = –h – k/2 and y1 = –3k + 2k – 8h = –k – 8h  16h2 + k2 + 10hk = 2. Hence locus of (h, k) is 16x2 + y2 + 10 xy = 2 Example 5 : Prove that if normal to the hyperbola xy = c2 at point t meets the curve again at a point t1 then t3 t1 + 1 = 0. Solution : Equationofnormalat point t i.e., (ct, c/t) is
• 3. y – xt2 = t c (1 – t4 ) ... (1) It meets the curve again at t1 then (ct1 , c/t1 ) must satisfy(1)  3 2 1 1 4 2 1 1 t t 1 t t t 1 ) t 1 ( t c t ct t c         0 ) t t ( t t 1 t 1 1 2 1      1 1 tt ) t t (  (1 + t3 t1 ) = 0 Clearly 1 t t   t3 t1 + 1 = 0. Example 6 : The angle betweena pair oftangents drawnfroma point P to the parabola y2 = 4ax is45°. Show that the locusofthe point P is a hyperbola. [IIT-1998] Solution : Let P (  , ) be anypoint on the locus. Equation ofpair oftangents fromP(  , ) to the parabola y2 = 4ax is [ ) ax 4 y ( ) a 4 ( ] ) x ( a 2 y 2 2 2          [T2 = SS1 ] ... (i) A = coefficient ofx2 = 4a2 2H = coefficient ofxy= –4 and B = coefficient of y2 =  2 – ( 2 – 4a ) = 4a . Since the angle between the two lines of(1) is 45°, we have 1 = tan45° = 2 2 H AB A B    (A + B)2 = 4 (H2 –AB)  (4a2 + 4  a )2 = 4[a2 2  – (4a)2 (4a )]  0 a a 6 2 2 2        or 2 2 2 a 8 ) a 3 (      The equation of required locus is (x + 3a)2 – y2 = 8a2 which is a hyperbola. Alternate Solution Equation ofanytangent to hyperbola y2 = 4ax is y = mx + a/m which passes through ( ,  )if  =  m + a/m or m2  – m + a = 0 ... (1) If m1 and m2 are roots of (1). m1 + m2 =  / and m1 m2 =  / a we have 1 = tan45° = 2 1 2 1 m m 1 m m    (1 + m1 m2 )2 = (m1 – m2 )2  (1 + m1 m2 )2 = (m1 + m2 )2 – 4m1 m2  (1 + a/ )2 = 2 ) / (   – 4a/  ( + a)2 =  2 – 4a or ( + 3a)2 – 2  = 8a2 The required locus is (x + 3a)2 –y2 = 8a2 which is a hyperbola.
• 4. Example 7 : Find the centre, eccentricity, foci, directories and the lengths ofthe transverseand conjugate axes of the hyperbola, whose equation is (x – 1)2 –2 (y – 2)2 + 6 = 0 Solution : The equation ofthe hyperbola can be written as (x – 1)2 – 2(y–2)2 + 6 = 0 or     1 3 ) 2 y ( 6 ) 1 x ( – 2 2 2 2     or     2 2 2 2 Y x 1 3 6   Where Y= (y–2) and x = (x–1) ... (1)  centre: X = 0, Y = 0 i.e., (x – 1) = 0, x = 1 & (y – 2) = 0, y = 2. So a = 3 and b = 6 so transverse axis = 3 2 , and conjugate axis = 6 2 . Also b2 = a2 (e2 – 1)  6 = 3 (e2 – 1) i.e., e = 3 In (X, Y) coordinates, fociare (0,  ae) i.e., (0,  3).  foci are (1, 2  3) i.e., (1, 5) and (1, –1) Equations ofdirectories,Y=  a/e.  directrices y – 2 =  1 3 / 3   or y = 3, y = 1 Example 8 : Find the equationand angle between the asymptotes ofthe hyperbola x2 + 2xy – 3y2 + x + 7y + 9 = 0 Solution : Let the combined equation of asymptotes x2 + 2xy – 3y2 + x + 7y +  = 0 Ifit representspair ofstraight lines abc + 2fgh – af 2 – bg2 – ch2 = 0  = –23/16  Asymptotes x2 + 2xy – 3y2 + 7y – 23/16 = 0 Required angle =  tan–1 2. Example 9 : Prove that thelocus ofa point whose chord ofcontact touches the circle inscribed on the straight line joining the foci of the hyperbola x2 /a2 – y2 / b2 = 1 as diameter is x2 /a4 + y2 / b4 = 1/(a2 + b2 ). Solution : Circle on the join of foci (ae, 0) and (–ae, 0) diameter is (x – ae) (x + ae) + (y – 0) (y – 0) = 0 i.e., x2 + y2 = a2 e2 = a2 + b2 ... (i) [ a2 e2 = a2 + b2 ] Let chord ofcontact ofP (x1 , y1 ) touchthe cirlce (i) Equation ofchord ofcontact ofP is [T = 0] xx1 /a2 – yy1 /b2 = 1 i.e., b2 x1 x–a2 y1 y – a2 b2 = 0 ... (ii)  ) b a ( ) y a x b ( b a 2 2 2 1 4 2 1 4 2 2     Hence locus of P (x1 , y1 ) is (b4 x2 + a4 y2 ) (a2 + b2 ) = a4 b4 .
• 5. Example 10 : An ellipse has eccentricity 2 1 and one focus at the point       1 , 2 1 P . Its one directrix is common tangent to the circle x2 + y2 = 1 and the hyperbola x2 – y2 = 1, nearer to P. The equation of the ellipse inthe standard form. Solution : The circle x2 + y2 = 1 and the hyperbola x2 – y2 = 1 touch eachother at the points ( 1, 0)and the common tangent at these point are x =  1. Since x = 1 is nearer to the focus       1 , 2 1 P , this is the directrix ofthe required ellipse. Therefore, the major axis is parallel to the axis passing through the focus       1 , 2 1 P . Hence the equation ofthe major axis is y= 1. Let a bethelengthofthesemimajor axis ofthe ellipse and let thecoordinates ofthe centreC ofthe ellipse be ) 1 , ( . Then CP =   2 1 = a.e = a × 2 1 ... (i)        2 1 e  and the distance ofthe directrix fromthe centre = e a .  1 –  = a 2 e a  ... (ii) x – y = 1 2 2 x = 1 x + y = 1 2 2 From(i) and (ii) we get a = 1 1 ande 3 2  . Ifb is the length of the semi minor axis ofthe ellipse, then b2 = a2 (1 – e2 )  b2 = 12 1 4 1 1 9 1         Hence the required equation oftheellipse is 1 12 1 ) 1 y ( 3 1 3 1 x 2 2 2                 or 2 3 1 x 9        + 12(y – 1)2 = 1
• 6. SOLVED OBJECTIVE EXAMPLES Example 1 : The equation ofa line passing throughthe centre ofarectangular hyperbola is x– y– 1 =0. Ifone ofits asymptotes is 3x – 4y– 6 = 0, the equation of the other asymptote is (A) 4x – 3y + 17 = 0 (B) –4x – 3y + 17 = 0 (C) –4x + 3y + 1 = 0 (D) 4x + 3y + 17 = 0 Solution : Weknowthat asymptotesofrectangularhyperbolaaremutuallyperpendicular,thusotherasymptote should be 4x+ 3y+  = 0. Intersection point ofasymptotes is also the centre ofthe hyperbola. Hence intersection point of4x + 3y +  = 0 and 3x – 4y– 6 = 0 should lie on the line x–y–1 = 0, using it  canbe easilyobtained. Hence (D) is the correct answer. Example 2 : The locus ofthe middle points ofchords of hyperbola 3x2 – 2y2 + 4x – 6y= 0 parallelto y= 2x is (A) 3x – 4y = 4 (B) 3x – 4y + 4 = 0 (C) 4x – 4y = 3 (D) 3x – 4y = 2 Solution : Let the mid point be (h, k). Equation ofa chord whose mid point is (h, k) would be T = S1 or 3x h – 2yk + 2(x + h) – 3(y+k) = 3h2 – 2k2 + 4h – 6k  x (3h + 2) –y (2k + 3) – (2h + 3k) – 3h2 + 2k2 = 0 Its slope is 3 k 2 2 h 3   = 2 (given)  3h = 4k + 4  Required locus is 3x – 4y= 4 Hence (A) is the correct answer. Example 3 : The tangent at a point P on the hyperbola 2 2 2 2 b y a x  =1 meets one of the directrix in F. If PF subtends an angle  at the corresponding focus, then  equals (A) 4 /  (B) 2 /  (C) 4 / 3 (D)  Solution : Let directrix be x= a/e and focus be S(ae, 0). Let P (a  sec , b  tan ) be any point onthe curve. Equation of tangent at P is b tan y a sec x    = 1. Let F be the intersection point of tangent of directrix, then F =          tan e ) e (sec b , e / a
• 7.  SF PS 2 b(sec e) b tan m ,m etan (a 1) a(sec e)           mSF .mPS = –1 Hence (B) is the correct answer. Example 4 : The line lx + my+ n = 0 will be a normal to the hyperbola b2 x2 – a2 y2 = a2 b2 if (A) 2 2 2 2 2 2 2 2 n ) b a ( m b a    l (B) 2 2 2 2 2 2 2 2 a b (a b ) m n    l (C) n ) b a ( m b a 2 2 2 2 2 2 2    l (D) noneofthese Solution : Equation ofnormalat (a sec, btan) is ax cos + by cot  = a2 + b2 Comparing it with lx + my + n = 0 we get l  cos a = 2 2 bcot (a b ) m n      cos = nb ) b a ( m cot and an ) b a ( 2 2 2 2       l  sin = am bl Thus 2 2 2 2 m a b l + 1 n a ) b a ( 2 2 2 2 2 2   l or, 2 2 2 2 2 2 2 2 n ) b a ( m b a    l Hence (B) is the correct answer. Example 5 : If (a sec, btan) and (asec , btan ) be the coordinate of the ends of a focal chord of 2 2 2 2 b y a x  = 1, then tan 2 tan 2   equals to (A) 1 e 1 e   (B) e 1 e 1   (C) c 1 e 1   (D) 1 e 1 e   Solution : Equation ofchord connecting the points (asec, b tan) and (asec, b  tan ) is
• 8. x y cos sin cos a 2 b 2 2                            Ifit passes through (ae, 0);we have, ecos cos 2 2                    e = cos 1 tan . tan 2 2 2 1 tan tan cos 2 2 2                           tan e 1 e 1 2 tan . 2      Hence (B) is the correct answer. Example 6 : The point ofintersectionofthe curves whose parametric equations arex= t3 + 1, y= 2t andx= 2s, y = 2 s , is given by (A) (1, –3) (B) (2, 2) (C) (–2, 4) (D) (1, 2) Solution : x = t2 + 1, y = 2t  x – 1 = y2 /2 x = 2s, y = 2/s  xy = 4 For the point ofintersection we have 0 16 y 4 y 4 y 1 y 4 3 2        y= 2  x = 2 Hence (B) is the correct answer. Example 7 : The equation ofthe hyperbola whose fociare (6, 5), (–4, 5) and eccentricity 4 5 is (A) 1 9 ) 5 y ( 16 ) 1 x ( 2 2     (B) 1 9 y 16 x 2 2   (C) 1 9 ) 5 y ( 16 ) 1 x ( 2 2      (D) None ofthese Solution : S1  (6, 5); S2 (–4, 5), e = 5/4 S1 S2 = 10  2ae = 10  a = 4 and b2 = a2 (e2 – 1) = 16 9 1 16 25         Centre of the hyperbola is (1, 5)  Equation ofrequiredhyperbola is 1 9 ) 5 y ( 16 ) 1 x ( 2 2    
• 9. Example 8 : The equation (x –  )2 + (y –  )2 = k(lx + my + n)2 represents (A) a parabola for k < (l2 + m2 )–1 (B) an ellipse for 0 < k < (l2 + m2 )–1 (C) a hyperbola for k > (l2 + m2 )–1 (D) a point circle for k = 0. Solution : (x –  )2 + 2 ) (    = k (lx + my + n)2 = k (l2 + m2 ) 2 2 2 m n my x            l l  PS k PM  (l2 + m2 ). If k(l2 + m2 ) = 1, P lies on parabola If k(l2 + m2 ) < 1, P lies on ellipse If k(l2 + m2 ) > 1, P lies on hyperbola Ifk = 0, P lies on a point circle Hence (B), (C), (D) are correct. Example 9 : The point on the hyperbola 1 18 y 24 x 2 2   which is nearest to the line 3x + 2y+ 1 = 0 is (A) (6, 3) (B) (–6, 3) (C) (6, –3) (D) (–6, –3) Solution : Equation oftangent is ) tan 18 , sec 24 (   is 1 18 tan y 24 sec x     , then point is nearest to the line 3x + 2y + 1 = 0. so its slope = – 2 3  2 3 tan 18 24 sec      sin = – 3 1 Hence the point is (6, –3) Hence (C) is the correct answer. Example 10 : The locus ofa point, fromwhere tangents to the rectangularhyperbola x2 – y2 = a2 containanangle of45° is (A) (x2 + y2 ) + a2 (x2 – y2 ) = 4a4 (B) 2(x2 + y2 ) + 4a2 (x2 – y2 ) = 4a2 (C) (x2 + y2 ) + 4a2 (x2 – y2 ) = 4a4 (D) (x2 + y2 ) + a2 (x2 – y2 ) = a4 Solution : Let y= mx 2 2 2 a a m   be two tangent and passes through (h, k) then (k – mk)2 = m2 a2 – a2  m2 (h2 – a2 ) – 2khm + k2 + a2 = 0  m1 + m2 = 2 2 a h kh 2  and m1 m2 = 2 2 2 2 a h a k   , using tan45°= 1 2 1 2 m m 1 m m   Hence (C) is the correct answer.
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