Maths 1B Important Questions<\/a> TS Inter First Year Maths 1B Locus Important Questions to help strengthen their preparations for exams.<\/p>\nTS Inter First Year Maths 1B Locus Important Questions<\/h2>\n
Question 1.
\nFind the equation of the locus of a point that is at a distance 5 from (-2, 3), in the XOY plane. [Mar. ’02; May ’95]
\nSolution:
\nLet A (-2, 3) be the given point.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA = 5
\n
\nSquaring on both sides
\n\u21d2 13 + x2<\/sup> + y2<\/sup> + 4x – 6y = 25
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 6y = 25 – 13
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 6y = 12
\n\u2234 The equation of locus of P(x, y) is x2<\/sup> + y2<\/sup> + 4x – 6y = 12.<\/p>\nQuestion 2.
\nFind the equation of the locus of a point P such that the distance of P from the origin is twice the distance of P from A(1, 2). [Mar. ’12; May ’05, ’97]
\nSolution:
\nO(0, 0), A(1, 2) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is OP = 2PA
\n
\n\u21d2 x2<\/sup> + y2<\/sup> = 4[(x – 1)2<\/sup> + (y – 2)2<\/sup>]
\n\u21d2 x2<\/sup> + y2<\/sup> = 4(x2<\/sup> – 2x + 1 + y2<\/sup> – 4y + 4)
\n\u21d2 x2<\/sup> + y2<\/sup> = 4x2<\/sup> – 8x + 4 + 4y2<\/sup> – 16y + 16
\n\u21d2 3x2<\/sup> + 3y2<\/sup> – 8x – 16y + 20 = 0
\n\u2234 The equation of the locus of P(x, y) is 3x2<\/sup> + 3y2<\/sup> – 8x – 16y + 20 = 0.<\/p>\n<\/p>\n
Question 3.
\nFind the equation of the locus of a point equidistant from A(2, 0) and the Y-axis. (May ’03)
\nSolution:
\nA(2, 0) is the given point.
\nLet P(x, y) be any point on the locus.
\n
\nThe distance of P from Y-axis = PN = |x|
\nThe given geometric condition is PA = |PN|
\n\\(\\sqrt{(x-2)^2+(y-0)^2}=|x|\\)
\nSquaring on both sides
\n\u21d2 \\(\\left(\\sqrt{(x-2)^2+(y-0)^2}\\right)^2=(|x|)^2\\)
\n\u21d2 (x – 2)2<\/sup> + (y)2<\/sup> = x2<\/sup>
\n\u21d2 x2<\/sup> – 4x + 4 + y2<\/sup> = x2<\/sup>
\n\u21d2 y2<\/sup> – 4x + 4 = 0
\n\u2234 The equation of the locus of P(x, y) is y2<\/sup> – 4x + 4 = 0.<\/p>\nQuestion 4.
\nFind the equation of the locus of a point P, the square whose distance from the origin is 4 times its y-coordinate. [Mar. ’00]
\nSolution:
\nO(0, 0) is the origin.
\nLet P(x, y) be any point on the locus.
\nThe distance of P from the X-axis = y.
\n
\nThe given geometric condition is OP2<\/sup> = 4 PM
\n\u21d2 \\(\\left(\\sqrt{(x-0)^2+(y-0)^2}\\right)^2=4 y\\)
\n\u21d2 x2<\/sup> + y2<\/sup> = 4y
\n\u21d2 x2<\/sup> + y2<\/sup> – 4y = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> – 4y = 0.<\/p>\nQuestion 5.
\nFind the equation of locus of a point P such that PA2<\/sup> + PB2<\/sup> = 2c2<\/sup>, where A = (a, 0), B = (-a, 0), and 0 < |a| < |c|. [Mar. ’00]
\nSolution:
\nA = (a, 0), B = (-a, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA2<\/sup> + PB2<\/sup> = 2c2<\/sup>
\n\u21d2 \\(\\left[\\sqrt{(x-a)^2+(y-0)^2}\\right]^2\\) + \\(\\left[\\sqrt{(x+a)^2+(y-0)^2}\\right]^2\\) = 2c2<\/sup>
\n\u21d2 (x – a)2<\/sup> + y2<\/sup> + (x + a)2<\/sup> + y2<\/sup> = 2c2<\/sup>
\n\u21d2 x2<\/sup> – 2ax + a2<\/sup> + y2<\/sup> + x2<\/sup> + 2ax + a2<\/sup> + y2<\/sup> = 2c2<\/sup>
\n\u21d2 2x2<\/sup> + 2y2<\/sup> + 2a2<\/sup> = 2c2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + a2<\/sup> = c2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = c2<\/sup> – a2<\/sup>
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> = c2<\/sup> – a2<\/sup><\/p>\nQuestion 6.
\nFind the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P. [Mar. ’13, ’05, ’03; May ’12, ’04, ’03, ’02]
\nSolution:
\nLet A(2, 3), B(-1, 5) are the given points.
\nLet P(x, y) be any point on the locus
\nThe given geometric condition is \u2220APB = 90\u00b0
\n
\nThen, PA2<\/sup> + PB2<\/sup> = AB2<\/sup>
\n\u21d2 \\(\\left[\\sqrt{(x-2)^2+(y-3)^2}\\right]^2+\\left[\\sqrt{(x+1)^2+(y-5)^2}\\right]^2\\) = \\(\\left[\\sqrt{(2+1)^2+(3-5)^2}\\right]^2\\)
\n\u21d2 (x – 2)2<\/sup> + (y – 3)2<\/sup> + (x + 1)2<\/sup> + (y – 5)2<\/sup> = (2 + 1)2<\/sup> + (3 – 5)2<\/sup>
\n\u21d2 x2<\/sup> + 4 – 4x + y2<\/sup> + 9 – 6y + x2<\/sup> + 1 + 2x + y2<\/sup> + 25 – 10y = 9 + 4
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 2x – 16y + 26 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – x – 8y + 13 = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> – x – 8y + 13 = 0.<\/p>\n<\/p>\n
Question 7.
\nFind the equation of the locus of P, if A = (4, 0), B = (-4, 0) and |PA – PB| = 4. [May ’13]
\nSolution:
\nA = (4, 0), B = (-4, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is |PA – PB| = 4
\n\u21d2 PA – PB = \u00b14
\n\u21d2 PA = \u00b14 + PB
\nSquaring on both sides
\n\u21d2 (PA)2<\/sup> = (\u00b14 + PB)2<\/sup>
\n\u21d2 PA2<\/sup> = 16 + PB2<\/sup> \u00b1 8PB
\n\u21d2 (x – 4)2<\/sup> + (y – 0)2<\/sup> = 16 + (x + 4)2<\/sup> + (y – 0)2<\/sup> \u00b1 8PB
\n\u21d2 x2<\/sup> – 8x + 16 + y2<\/sup> = 16 + x2<\/sup> + 8x + 16y2<\/sup> \u00b1 8PB
\n\u21d2 -8x = 8x + 16 \u00b1 8PB
\n\u21d2 -8x – 8x – 16 = \u00b18PB
\n\u21d2 -16 – 16x = \u00b18PB
\n\u21d2 -2 – 2x = \u00b1PB
\nSquaring on both sides
\n\u21d2 (-2 – 2x)2<\/sup> = (\u00b1PB)2<\/sup>
\n\u21d2 4 + 8x + 4x2<\/sup> = PB2<\/sup>
\n\u21d2 4 + 4x2<\/sup> + 8x = (x + 4)2<\/sup> + (y – 0)2<\/sup>
\n\u21d2 4 + 4x2<\/sup> + 8x = x2<\/sup> + 8x + 16 + y2<\/sup>
\n\u21d2 3x2<\/sup> – y2<\/sup> – 12 = 0
\n\u2234 The equation of the locus of P(x, y) is 3x2<\/sup> – y2<\/sup> – 12 = 0.<\/p>\nQuestion 8.
\nFind the equation of the locus of a point, the sum of whose distances from (0, 2) and (0, -2) is 6. [Mar. ’16 (TS)]
\nSolution:
\nLet A = (0, 2), B = (0, -2) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA + PB = 6
\n\u21d2 PA = 6 – PB
\nSquaring on both sides
\n\u21d2 PA2<\/sup> = (6 – PB)2<\/sup>
\n\u21d2 PA2<\/sup> = 36 – 12PB + PB2<\/sup>
\n\u21d2 (x – 0)2<\/sup> + (y – 2)2<\/sup> = 36 + (x – 0)2<\/sup> + (y + 2)2<\/sup> – 12PB
\n\u21d2 x2<\/sup> + y2<\/sup> + 4 – 4y = 36 + x2<\/sup> + y2<\/sup> + 4 + 4y – 12PB
\n\u21d2 -4y = 36 + 4y – 12PB
\n\u21d2 12PB = 36 + 4y + 4y
\n\u21d2 12PB = 36 + 8y
\n\u21d2 3PB = 9 + 2y
\nSquaring on both sides
\n\u21d2 (3PB)2<\/sup> = (9 + 2y)2<\/sup>
\n\u21d2 9PB2<\/sup> = (9 + 2y)2<\/sup>
\n\u21d2 9[(x – 0)2<\/sup> + (y + 2)2<\/sup>] = 81 + 36y + 4y2<\/sup>
\n\u21d2 9[x2<\/sup> + y2<\/sup> + 4y + 4] = 81 + 36y + 4y2<\/sup>
\n\u21d2 9x2<\/sup> + 9y2<\/sup> + 36y + 36 = 81 + 36y + 4y2<\/sup>
\n\u21d2 9x2<\/sup> + 9y2<\/sup> + 36 – 81 – 4y2<\/sup> = 0
\n\u21d2 9x2<\/sup> + 5y2<\/sup> – 45 = 0
\n\u2234 The equation of the locus of P(x, y) is 9x2<\/sup> + 5y2<\/sup> – 45 = 0.<\/p>\nQuestion 9.
\nA(2, 3) and B(-3, 4) are two given points. Find the equation of the locus of P, so that the area of the triangle PAB is 8.5. [Mar. ’11]
\nSolution:
\nA(2, 3), B(-3, 4) are the points given.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is an area of the triangle, PAB = 8.5
\n
\n\u2234 The equation of the locus of P(x, y) is (x + 5y) (x + 5y – 34) = 0
\n\u21d2 x2<\/sup> + 5xy – 34x + 5xy + 25y2<\/sup> – 170y = 0
\n\u21d2 x2<\/sup> + 10xy + 25y2<\/sup> – 34x – 170y = 0<\/p>\n<\/p>\n
Question 10.
\nFind the equation of the locus of p, if the ratio of the distances from p to A(5, -4) and B(7, 6) is 2 : 3. [Mar. ’14, ’98; May ’08, ’01]
\nSolution:
\nLet A(5, -4), B(7, 6) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA : PB = 2 : 3
\n\u21d2 \\(\\frac{\\mathrm{PA}}{\\mathrm{PB}}=\\frac{2}{3}\\)
\n\u21d2 3PA = 2PB
\n
\n\u21d2 9[x2<\/sup> – 10x + 25 + y2<\/sup> + 16 + 8y] = 4[x2<\/sup> – 14x + 49 + y2<\/sup> – 12y + 36]
\n\u21d2 9x2<\/sup> – 90x + 225 + 9y2<\/sup> + 144 + 72y = 4x2<\/sup> – 56x + 196 – 4y2<\/sup> – 48y + 144
\n\u21d2 9x2<\/sup> – 90x + 225 + 9y2<\/sup> + 144 + 72y – 4x2<\/sup> + 56x – 196 – 4y2<\/sup> + 48y – 144 = 0
\n\u21d2 5x2<\/sup> + 5y2<\/sup> – 34x + 120y + 29 = 0
\n\u2234 The equation of the locus of P(x, y) is 5x2<\/sup> + 5y2<\/sup> – 34x + 120y + 29 = 0<\/p>\nQuestion 11.
\nA(1, 2), B(2, -3) and C(-2, 3) are three points. A point p moves such that PA2<\/sup> + PB2<\/sup> = 2PC2<\/sup>. Show that the equation to the locus of P is 7x – 7y + 4 = 0. [Mar. ’19 (T.S); Mar. ’17 (AP); May ’15 (A.P); ’07]
\nSolution:
\nA(1, 2), B(2, -3), C(-2, 3) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA2<\/sup> + PB2<\/sup> = 2PC2<\/sup>
\n\u21d2 [(x – 1)2<\/sup> + (y – 2)2<\/sup>] + [(x – 2)2<\/sup> + (y + 3)2<\/sup>] = 2[(x + 2)2<\/sup> + (y – 3)2<\/sup>]
\n\u21d2 x2<\/sup> – 2x + 1 + y2<\/sup> + 4 – 4y + x2<\/sup> + 4 – 4x + y2<\/sup> + 9 + 6y = 2[x2<\/sup> + 4 + 4x + y2<\/sup> + 9 – 6y]
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 6x + 2y + 18 = 2x2<\/sup> + 8 + 8x + 2y2<\/sup> + 18 – 12y
\n\u21d2 -6x + 2y = 8 + 8x – 12y
\n\u21d2 8 + 8x – 12y + 6x – 2y = 0
\n\u21d2 14x – 14y + 8 = 0
\n\u21d2 7x – 7y + 4 = 0
\n\u2234 The equation of the locus of P(x, y) is 7x – 7y + 4 = 0.<\/p>\nSome More Maths 1B Locus Important Questions<\/h3>\n
Question 12.
\nFind the equation of the locus of a point that is at a distance 5 from A(4, -3).
\nSolution:
\nA(4, -3) is the given point.
\nLet P(x, y) be any point on the locus.
\n
\nThe given geometric condition is PA = 5
\n\u21d2 \\(\\sqrt{(x-4)^2+(y+3)^2}\\) = 5
\nSquaring on both sides
\n\u21d2 \\(\\left(\\sqrt{(x-4)^2+(y+3)^2}\\right)^2\\) = (5)2<\/sup>
\n\u21d2 (x – 4)2<\/sup> + (y + 3)2<\/sup> = 25
\n\u21d2 x2<\/sup> – 8x + 16 + y2 + 9 + 6y = 25
\n\u21d2 x2<\/sup> + y2 – 8x + 6y = 0
\n\u2234 The equation of locus of P(x, y) is x2<\/sup> + y2<\/sup> – 8x + 6y = 0<\/p>\nQuestion 13.
\nFind the equation of the locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).
\nSolution:
\nA(3, 0), B(-3, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA = 2PB
\n
\n\u21d2 (x – 3)2<\/sup> + (y – 0)2<\/sup> = 4[(x + 3)2<\/sup> + (y – 0)2<\/sup>]
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup>\u00a0<\/sup>= 4[x2<\/sup> + 6x + 9 + y2<\/sup>]
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup> = 4×2 + 24x + 36 + 4y2<\/sup>
\n\u21d2 4x2<\/sup> + 24x + 36 + 4y2 – x2<\/sup> + 6x – 9 – y2<\/sup> = 0
\n\u21d2 3x2<\/sup> + 3y2<\/sup> + 30x + 27 = 0
\n\u21d2 x2<\/sup> + y2<\/sup>\u00a0+ 10x + 9 = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> + 10x + 9 = 0.<\/p>\n<\/p>\n
Question 14.
\nThe ends of the hypotenuse of a right-angled triangle are (0, 6) and (6, 0). Find the equation of the locus of its third vertex. [June ’10; Mar. ’08; Sept. ’00]
\nSolution:
\nLet A(0, 6), B(6, 0) are the given points.
\nLet P(x, y) be the third vertex
\n
\nThe given geometric condition is \u2220APB = 90\u00b0
\n\u21d2 PA2<\/sup> + PB2<\/sup> = AB2<\/sup>
\n\u21d2 \\(\\left[\\sqrt{(x-0)^2+(y-6)^2}\\right]^2+\\left[\\sqrt{(x-6)^2+(y-0)^2}\\right]^2\\) = \\(\\left[\\sqrt{(0-6)^2+(6-0)^2}\\right]^2\\)
\n\u21d2 x2<\/sup> + (y – 6)2<\/sup> + (x – 6)2<\/sup> + y2<\/sup> = (-6)2<\/sup> + (6)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> – 12y + 36 + x2<\/sup> – 12x + 36 + y2<\/sup> = 36 + 36
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 12x – 12y + 72 = 72
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 12x – 12y = 0
\n\u21d2 x2<\/sup> + y