{"id":35524,"date":"2022-09-19T17:21:48","date_gmt":"2022-09-19T11:51:48","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35524"},"modified":"2022-09-19T17:22:03","modified_gmt":"2022-09-19T11:52:03","slug":"inter-2nd-year-maths-2b-hyperbola-solutions-ex-5a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-hyperbola-solutions-ex-5a\/","title":{"rendered":"Inter 2nd Year Maths 2B Hyperbola Solutions Ex 5(a)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nOne focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is the line y = 2. Find the equation of the hyperbola if its eccentricity is \\(\\frac{3}{2}\\).
\nSolution:
\nS(1 – 3) is the focus, equation of the directrix is y – 2 = 0
\n\"Inter
\nP(x1<\/sub>, y1<\/sub>) is any point on the hyperbola join SP and draw PM perpendicular to the directrix S.P = e. PM \u21d2 SP\u00b2 = e\u00b2. PM\u00b2
\n(x1<\/sub> – 1)\u00b2 + (y1<\/sub> + 3)\u00b2 = \\(\\frac{9}{4}\\)|\\(\\frac{(y_1-2}{\\sqrt{1+0}}\\)|\u00b2
\nx\u00b21<\/sub> + 1 – 2x1<\/sub> + y\u00b21<\/sub> + 9 + 6y1<\/sub> = \\(\\frac{9}{4}\\) (y1<\/sub> – 2)\u00b2
\n4x\u00b21<\/sub> + 4y\u00b21<\/sub> – 8x1<\/sub> + 24y1<\/sub> + 40 = 9(y\u00b21<\/sub> + 4 – 4y1<\/sub>) = 9y\u00b21<\/sub> – 36y1<\/sub> + 36
\n4x\u00b21<\/sub> – 5y\u00b21<\/sub> – 8x1<\/sub> + 60y1<\/sub> + 4 = 0
\nFocus of P(x\u00b2, y\u00b2) is
\n4x\u00b2 – 5y\u00b2 – 8x + 60y + 4 = 0
\nThis is the equation of the required hyperbola.<\/p>\n

Question 2.
\nIf the lines 3x – 4y = 12 and 3x + 4y = 12 meets on a hyperbola S = 0 then find the eccentricity of the hyperbola S = 0.
\nSolution:
\nThe given lines as
\n3x – 4y = 12
\n3x + 4y = 12
\nThe combined equation of the lines
\n(3x – 4y) (3x + 4y) = 144
\n9x\u00b2 – 16y\u00b2 = 144
\n\"Inter<\/p>\n

Question 3.
\nFind the equations of the hyperbola whose foci are (\u00b15, 0) the transverse axis is of length 8.
\nSolution:
\nFoci are S(\u00b15, 0) \u2234 ae = 5
\nLength of transverse axis = 2a = 8
\na = 4
\ne = \\(\\frac{5}{4}\\)
\nb\u00b2 = a\u00b2 (e\u00b2 – 1) = 16(\\(\\frac{25}{16}\\) – 1)
\nEquation of the hyperbola is \\(\\frac{x^2}{16}-\\frac{y^2}{9}\\) = 1
\n9x\u00b2 – 16y\u00b2 = 144<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the equation of the hyperbola, whose asymptotes are the straight lines (x + 2y + 3) = 0, (3x + 4y + 5) = 0 and which passes through the point (1, -1).
\nSolution:
\nCombined equation of the asymptotes as
\n(x + 2y + 3) (3x + 4y + 5) = 0
\n\u2234 Equation of the hyperbola can be taken as
\n(x + 2y + 3) (3x + 4y + 5) + k = 0
\nThe hyperbola passes through P(1, -1)
\n(1 – 2 + 3) (3-4 + 5) + k = 0
\n8 + k = 0 \u21d2 k = -8
\nEquation of the hyperbola is
\n(x + 2y + 3) (3x + 4y + 5) – 8 = 0
\n3x\u00b2 + 6xy + 9x + 4xy + 8y\u00b2 + 12y + 5x +10y + 15 – 8 = 0
\n3x\u00b2 + 10xy + 8y\u00b2 + 14x + 22y + 7 = 0<\/p>\n

Question 5.
\nIf 3x – 4y + k = 0 is a tangent to x\u00b2 – 4y\u00b2 = 5, find value of k.
\nSolution:
\nEquation of the hyperbola x\u00b2 – 4y\u00b2 = 5
\n\\(\\frac{x^2}{5}-\\frac{y^2}{\\left(\\frac{5}{{4}}\\right)}\\) = 1
\na\u00b2 = 5, b\u00b2 = \\(\\frac{5}{4}\\)
\nEquation of the given line is 3x – 4y + k = 0
\n4y = 3x + k
\ny = \\(\\frac{3}{4}\\)x + \\(\\frac{k}{4}\\)
\nm = \\(\\frac{3}{4}\\) c = \\(\\frac{k}{4}\\)
\nCondition for tangency is c\u00b2 = a\u00b2m\u00b2 – b\u00b2
\n\\(\\frac{k^2}{16}\\) = 5. \\(\\frac{9}{16}-\\frac{5}{4}\\)
\nk\u00b2 = 45 – 20 = 25
\nk = \u00b15.<\/p>\n

Question 6.
\nFind the product of lengths of the perpen-diculars from any point on the hyperbola \\(\\frac{x^2}{16}-\\frac{y^2}{9}\\) = 1 to its asymptotes.
\nSolution:
\nEquation of the hyperbola is \\(\\frac{x^2}{16}-\\frac{y^2}{9}\\) = 1
\na\u00b2 = 16, b\u00b2 = 9
\nProduct of the perpendiculars from any point as the hyperbola to its asymptotes
\n\"Inter<\/p>\n

Question 7.
\nIf the eccentricity of a hyperbola is \\(\\frac{5}{4}\\), then find the eccentricity of its conjugate-hyperbola.
\nSolution:
\nIf e and er, are the eccentricity of a hyperbola and its conjugate hyperbola, then
\n\"Inter<\/p>\n

Question 8.
\nFind the equation of the hyperbola whose asymptotes are 3x = \u00b1 5y and the vertices are (\u00b1 5, 0).
\nSolution:
\nThe equation of the asymptotes are
\n3x = \u00b15y
\n3x – 5y = 0, 3x + 5y = 0
\nCombined equation of the asymptotes is
\n(3x – 5y) (3x + 5y) = 0
\n9x\u00b2 – 25y\u00b2 = 0
\nEquation of the hyperbola is 9x\u00b2 – 25y\u00b2 = k
\nThe hyperbola passes through the vertex (5, 0)
\n9(5)\u00b2 – 0 = k \u21d2 k = 225
\nEquation of the hyperbola is
\n9x\u00b2 – 25y\u00b2 = 225<\/p>\n

\"Inter<\/p>\n

Question 9.
\nFind the equation of the normal at \u03b8 = \\(\\frac{\\pi}{3}\\) to the hyperbola 3x\u00b2 – 4y\u00b2 = 12.
\nSolution:
\nEquation of the hyperbola is 3x\u00b2 – 4y\u00b2 = 12
\n\\(\\frac{x^2}{4}+\\frac{y^2}{3}\\) = 1
\nEquation of the normal is
\n\"Inter<\/p>\n

Question 10.
\nIf the angle between the asymptotes is 30\u00b0 then find its eccentricity.
\nSolution:
\nAngle between the asymptotes = 2\u03b8 = 30\u00b0
\n\u03b8 = 15\u00b0
\n\"Inter
\nEccentricity = e = \u221a6 – \u221a2<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the following hyperbola.
\ni) 16y\u00b2 – 9x\u00b2 = 144
\nSolution:
\nEquation of the hyperbola is
\n16y\u00b2 – 9x\u00b2 = 144
\n\\(\\frac{y^2}{9}-\\frac{x^2}{16}\\) = 1
\na\u00b2 = 16, b\u00b2 = 9
\nCentre C(0, 0)
\nb\u00b2e\u00b2 = a\u00b2 + b\u00b2 = 16 + 9
\n= 25 \u21d2 be = 5
\nFoci are 5(0, \u00b1ae) = (0, \u00b15)
\nEccentricity = \\(\\frac{be}{b}=\\frac{5}{3}\\)
\nEquation of the directrices are y = \u00b1b\/e
\n\u00b1 3. \\(\\frac{5}{3}\\)
\n5y = \u00b1 9
\nLength of the latus return = 2. \\(\\frac{a^2}{b}\\)
\n= 2. \\(\\frac{16}{3}\\) = \\(\\frac{32}{3}\\)<\/p>\n

ii) x\u00b2 – 4y\u00b2 = 4
\nSolution:
\nEquation of the hyperbola is \\(\\frac{x^2}{4}-\\frac{y^2}{1}\\) = 1
\na\u00b2 = 4, b\u00b2 = 1
\nCentre is c (0,0)
\na\u00b2e\u00b2 = a\u00b2 + b\u00b2 = 4 + 1 = 5
\nae = \u221a5
\nFoci are (\u00b1ae, 0) = (\u00b1 \u221a5, 0)
\nEccentricity = \\(\\frac{ae}{a}=\\frac{\\sqrt{5}}{2}\\)
\nEquations of directrices are x = \u00b1\\(\\frac{a}{e}\\)
\n= \u00b1 2. \\(\\frac{2}{\\sqrt{5}}\\)
\n\u21d2 \u221a5 = \u00b1 4
\n\u21d2 \u221a5x \u00b1 4 = 0
\nLength of the latus rectum = \\(\\frac{2b^2}{a}=\\frac{2.1}{2}\\) = 1<\/p>\n

iii) 5x\u00b2 – 4y\u00b2 + 20x + 8y = 4
\nSolution:
\n5(x\u00b2 + 4x + 4) -4 (y\u00b2 – 2y + 1) = 4 + 20 – 4
\n5(x + 2)\u00b2 – 4(y – 1)\u00b2 = 20
\n\\(\\frac{(x+2)^2}{4}+\\frac{(y-1)^2}{5}\\) = 1
\na\u00b2 = 4, b\u00b2 = 5 => a < b
\nCentre C(-2, +1)
\na\u00b2e\u00b2 = a\u00b2 + b\u00b2 = 4 + 5 = 9
\nae = 3
\nEccentricity = \\(\\frac{ae}{a}=\\frac{3}{2}\\)
\nFoci are (h \u00b1 ae, k) = (-2 \u00b1 3, 1)
\n= (-5, 1) and (1, 1)
\nEquations of the directrices are x – h = \u00b1\\(\\frac{a}{e}\\)
\nx + 2 = \u00b1 2. \\(\\frac{2}{3}\\)
\n3x + 6 = \u00b1 4
\n3x+ 10 = 0 or 3x + 2 = 0
\nLength of the latus rectum = \\(\\frac{2b^2}{a}=\\frac{2.5}{2}\\) = 5<\/p>\n

iv) 9x\u00b2 – 16y\u00b2 + 72x – 32y – 16 = 0
\nSolution:
\nEquation of the hyperbola is
\n9x\u00b2 – 16y\u00b2 + 72x – 32y -16 = 0
\n\u21d2 9(x\u00b2 +8x) – 16(y\u00b2 + 2y) = 16
\n\u21d2 9(x\u00b2 + 8x +16) – 16 (y\u00b2 + 2y + 1)
\n= 16 + 144 – 16
\n\u21d2 9(x + 4)\u00b2 – 16(y + 1)\u00b2 = 144
\n\\(\\frac{(x+4)^2}{16}+\\frac{(y+1)^2}{9}\\) = 1
\nComparing with \\(\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}\\) = 1
\na\u00b2 = 16, b\u00b2 = 9, h = -4, k = -1
\nCentre (h, k) = (-4, -1)
\n\"Inter
\nFoci = (h \u00b1 ae, k) = (- 4\u00b14.\\(\\frac{5}{4}\\), 1)
\n= (-4 \u00b1 5, -1)
\n= (1, -1) and (-9, -1)
\nEquation of the directrices are
\nx + 4 = \u00b1 4. \\(\\frac{4}{5}\\)
\n= \u00b1 \\(\\frac{16}{5}\\)
\n5x + 20 = \u00b1 16
\nEquation of the directices are 5x + 4 = 0
\nand 5x + 36 = 0
\nLength of Latus rectum = \\(2\\frac{b^2}{a}=\\frac{2.9}{4}=\\frac{9}{2}\\)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the equation to the hyperbola whose foci are (4, 2) and (8, 2) and eccentricity is 2.
\nSolution:
\nFoci are (4, 2) and (8, 2) Centre C is the mid point of the foci
\n\u2234 Centre is (\\(\\frac{4+8}{2}\\), \\(\\frac{2+2}{2}\\)) = (6, 2)
\nae = 6 – 4 = 2
\nGiven e = 2 \u21d2 a = \\(\\frac{ae}{e}=\\frac{2}{2}\\) = 1
\nb\u00b2 = a\u00b2 (e\u00b2 – 1) = 1(4 – 1) = 3
\nEquation of the hyperbola is
\n\"Inter
\nMultiplying with 3
\n3(x – 6)\u00b2 – (y – 2)\u00b2 = 3
\n\u21d2 3(x\u00b2 – 12x + 36) – (y\u00b2 – 4y + 4) = 3
\n\u21d2 3x\u00b2 – 36x + 108 – y\u00b2 + 4y – 4 -3 = 0
\n3x\u00b2 – y\u00b2 – 36x + 4y + 101 = 0<\/p>\n

Question 3.
\nFind the equation of the hyperbola of given length of transverse axis 6 whose vertex bisects the distance between the centre and the focus.
\nSolution:
\nGiven CA = AS
\n\"Inter
\na = ae – a
\n2a = ae \u21d2 e = 2
\nLength of transverse axis = 2a = 6 \u21d2 a = 3
\nb\u00b2 = a\u00b2(e\u00b2 – 1) = 9(4 – 1) = 27
\nEquation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1
\n\\(\\frac{x^2}{9}-\\frac{y^2}{27}\\) = 1
\n3x\u00b2 – y\u00b2 = 27<\/p>\n

Question 4.
\nFind the equations of the tangents to the hyperbola x\u00b2 – 4y\u00b2 = 4 Which are i) Parallel ii) Perpendicular to the line x + 2y = 0.
\nSolution:
\nEquation of the hyperbola is x\u00b2 – 4y\u00b2 = 4
\n\\(\\frac{x^2}{4}-\\frac{y^2}{1}\\) = 1
\na\u00b2 = 4, b\u00b2 = 1<\/p>\n

i) The+anget is parallel to x + 2y = 0
\nm = –\\(\\frac{1}{2}\\)
\nc\u00b2 = a\u00b2m\u00b2 – b\u00b2 = 4.\\(\\frac{1}{4}\\) – 1 = 1 – 1 = 0
\nc = 0
\nEquation of the parallel tangent is
\ny = mx + c
\n= –\\(\\frac{1}{2}\\)x
\n2y = -x
\nx + 2y = 0.<\/p>\n

ii) The tangent is perpendicular to x +2y = 0
\nSlope of the tangent = m = \\(\\frac{-1}{\\left(-\\frac{1}{2}\\right)}\\) = 2
\nc\u00b2 = a\u00b2m\u00b2 – b\u00b2 = 4. 4 – 1 = 16
\nc = \u00b1\u221a15
\nEquation of the perpendicular tangent is
\ny = 2x \u00b1 \u221a15<\/p>\n

Question 5.
\nFind the equations of tangents drawn to the hyperbola 2x\u00b2 – 3y\u00b2 = 6 through (-2, 1).
\nSolution:
\nEquation of the hyperbola is 2x\u00b2 – 3y\u00b2 = 6
\n\\(\\frac{x^2}{3}-\\frac{y^2}{2}\\) = 1
\nSuppose the slope of the tangent is m\u00b9
\nThe tangent passes through P (-2, 1)
\nEquation of the tangent is
\ny – 1 = m(x + 2) = mx + 2m
\ny = mx + (2m + 1) ………. (1)
\nCondition for tangency is c\u00b2 = a\u00b2m\u00b2 – b\u00b2
\n(2m + 1)\u00b2 = 3m\u00b2 – 2
\n4m\u00b2 + 4m + 1 = 3m\u00b2 – 2
\nm\u00b2 + 4 m + 3 = 0
\n(m + 1) (m + 3) = 0
\nm = -1 or – 3<\/p>\n

Case (i): m = -1
\nSubstituting in (1) equation of the tangent is
\ny = -x – 1
\nx + y + 1 =0<\/p>\n

Case (ii) : m = – 3
\nEquation of the tangent is
\ny = – 3x – 5
\n3x + y + 5 = 0<\/p>\n

Question 6.
\nProve that the product of the perpendicular distance from any point on a hyperbola to its asymptotes is constant.
\nSolution:
\nEquation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1
\nAny point on the hyperbola is P(a sec \u03b8, b tan \u03b8)
\nEquations of the asymptotes are \\(\\frac{x}{a}\\) = \u00b1 \\(\\frac{y}{b}\\)
\ni.e., \\(\\frac{x}{a}-\\frac{y}{b}\\) = 0 and \\(\\frac{x}{a}+\\frac{y}{b}\\) = 0
\nPM = Perpendicular distance from P on
\n\"Inter
\nPN = Perpendicular distance from P on
\n\"Inter
\n\"Inter<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nTangents to the hyperbola make angles \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1 make angles \u03b81<\/sub>, \u03b82<\/sub> with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve 2xy = k(x\u00b2 – a\u00b2) when tan \u03b81<\/sub> + tan \u03b82<\/sub> = k.
\nSolution:
\nEquation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1
\nEquation of the tangent to the hyperbola can be taken as
\ny = mx \u00b1 \\(\\sqrt{a^2m^2-b^2}\\)
\nSuppose p(x1<\/sub>, y1<\/sub>) is the point of intersection of the tangents
\ny1<\/sub> = mx1<\/sub> \u00b1 \\(\\sqrt{a^2m^2-b^2}\\)
\ny1<\/sub> – mx1<\/sub> = \u00b1\\(\\sqrt{a^2m^2-b^2}\\)
\nSquaring both sides
\n(y1<\/sub> – mx1<\/sub>)\u00b2 = a\u00b2m\u00b2 – b\u00b2
\ny\u00b21<\/sub> + m\u00b2x\u00b21<\/sub> – 2mx1<\/sub>y1<\/sub> – a\u00b2m\u00b2 + b\u00b2 = 0
\nm\u00b2 (x\u00b21<\/sub> – a\u00b2) – 2mx1<\/sub>y1<\/sub> + (y\u00b21<\/sub> + b\u00b2) = 0
\nThis is a quadratic equation in m giving the values for m say m1<\/sub>, m2
\n\"Inter<\/sub>
\ni.e.,k = \\(\\frac{2x_1y_1}{x^2_1-a^2}\\)
\nor 2x1<\/sub>y1<\/sub> = k(x\u00b21<\/sub> – a\u00b21<\/sub>)
\np(x1<\/sub>, y1<\/sub>) lies on the curve 2xy = k(x\u00b2 – a\u00b2)<\/p>\n

Question 2.
\nShow that the locus of feet of the perpen-diculars drawn from foci to any tangent of the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1 is the auxiliary circle of the hyperbola.
\nSolution:
\nEquation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1
\nEquation of the tangent to the hyperbola is
\n\"Inter
\ny = mx \u00b1 \\(\\sqrt{a^2m^2-b^2}\\)
\ny – mx = \u00b1 \\(\\sqrt{a^2m^2-b^2}\\) ………… (1)
\nEquation to the perpendicular from foci (\u00b1ae, 0) on this tangent is 1
\ny = –\\(\\frac{1}{m}\\)(x \u00b1 ae)
\n\u21d2 my = – (x \u00b1 ae)
\nx + my = \u00b1ae …………. (2)
\nSquaring and adding (1) and (2)
\n(y – mx)\u00b2 + (x + my)\u00b2 = a\u00b2m\u00b2 – b\u00b2 + a\u00b2e\u00b2
\n\u21d2 y\u00b2 + m\u00b2x\u00b2 – 2mxy + x\u00b2 + m\u00b2y\u00b2 + 2mxy
\n= a\u00b2m\u00b2 – a\u00b2(e\u00b2 – 1) + a\u00b2e\u00b2 .
\n\u21d2 (x\u00b2 + y\u00b2) (1 + m\u00b2) = a\u00b2m\u00b2 – a\u00b2e\u00b2 + a\u00b2 + a\u00b2e\u00b2
\n= a\u00b2(1 + m\u00b2)
\nor x\u00b2 + y\u00b2 = a\u00b2 which is the auxiliary circle.
\nLocus of the feet of the perpendiculars drawn from the foci to any tangent to the hyperbola lies on the auxiliary circle.<\/p>\n

Question 3.
\nShow that the equation \\(\\frac{x^2}{9-c}-\\frac{y^2}{5-c}\\) = 1 represents.
\ni) An ellipse if ‘c’ is a real constant less than 5.
\nii) A hyperbola if ‘c’ is any real constant between 5 and 9.
\niii) Show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (\u00b12, 0) independent of the value of ‘c’.
\ni) An ellipse if ‘c’ is a real constant less than 5.
\nSolution:
\nGiven equation is
\n\\(\\frac{x^2}{9-c}-\\frac{y^2}{5-c}\\) = 1
\nThis equation represents an ellipse if 9 – c >0, 5 – c > 0
\n\u2234 c < 9, c < 5
\n\u21d2 c < 5<\/p>\n

ii) A hyperbola if ‘c’ is any real constant between 5 and 9.
\nSolution:
\nGiven equation is \\(\\frac{x^2}{9-c}-\\frac{y^2}{5-c}\\) = 1
\nThis equation represents a hyperbola if
\n9 – c > 0 and 5 – c < 0
\n9 > c and 5 < c
\nic. 5 < c < 9<\/p>\n

iii) Show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (\u00b12, 0), independent of the value ‘c’.
\nSolution:
\nIn both cases
\nCase (i): a\u00b2 = 9 – c, b\u00b2 = 5 – c
\na\u00b2 – b\u00b2 = (9 – c) – (5 – c)
\n= 9 – c – 5 + c = 4
\na\u00b2e\u00b2 = 4 \u21d2 ae = 2
\nFoci are (\u00b1ae, 0) (\u00b12, 0)<\/p>\n

Case (ii) : a\u00b2 = 9 – c, b\u00b2 = c – 5
\na\u00b2 + b\u00b2 = 9 – c + c – 5 = 4
\na\u00b2e\u00b2 = 4 \u21d2 ae = 2
\nFoci are (\u00b1 ae, 0) = (\u00b1 2, 0)<\/p>\n

\"Inter<\/p>\n

Question 4.
\nShow that the angle between two asymptotes of a hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}\\) = 1 is 2 Tan-1<\/sup>(\\(\\frac{b}{a}\\)) Or 2 Sec-1<\/sup>(e).
\nSolution:
\nEquations of the asymptotes are \\(\\frac{x}{a}-\\frac{y}{b}\\) = 0
\nand \\(\\frac{x}{a}+\\frac{y}{b}\\) = 0
\n\"Inter
\nIf 2 \u03b8 is the angle between the asymptotes then tan \u03b8 = \\(\\frac{b}{a}\\) = slope of the asymptotes of \u03b8 = tan-1<\/sup>\\(\\frac{b}{a}\\)
\nAngle between the asymptotes
\n\"Inter
\nAngle between the asymptotes
\n= 2tan-1<\/sup>\\(\\frac{b}{a}\\) or 2 Sec-1<\/sup>(e)<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a) I. Question 1. One focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35524"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35524"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35524\/revisions"}],"predecessor-version":[{"id":35526,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35524\/revisions\/35526"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35524"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35524"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35524"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}