{"id":3808,"date":"2023-12-16T10:28:28","date_gmt":"2023-12-16T04:58:28","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=3808"},"modified":"2023-12-18T17:53:54","modified_gmt":"2023-12-18T12:23:54","slug":"ts-inter-1st-year-maths-1b-solutions-chapter-1-ex-1a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1b-solutions-chapter-1-ex-1a\/","title":{"rendered":"TS Inter 1st Year Maths 1B Solutions Chapter 1 Locus Ex 1(a)"},"content":{"rendered":"

Students must practice these TS Intermediate Maths 1B Solutions<\/a> Chapter 1 Locus Ex 1(a) to find a better approach to solving the problems.<\/p>\n

TS Inter 1st Year Maths 1B Locus Solutions Exercise 1(a)<\/h2>\n

Question 1.
\nFind the equation of locus of a point which is at a distance 5 from A (4, – 3). (V.S.A.Q)
\nAnswer:
\nGiven A = (4, – 3) and suppose P (x, y) is any other point on the locus.
\nThen given PA = 5
\n\u21d2 PA2<\/sup> = 25
\n\u21d2 (x – 4)2<\/sup> + (y + 3)2<\/sup> = 25
\n\u21d2 x2<\/sup> + y2<\/sup> – 8x + 6y + 25 – 25 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 8x + 6y = 0 ……………… (1)
\n(If there exists another point Q(x1<\/sub>, y1<\/sub>) such that QA2<\/sup> = (x1<\/sub> – 4)2<\/sup> + (y1<\/sub> + 3)2<\/sup>
\nLet Q(x1<\/sub>, y1<\/sub> satisfy (1) then
\nx1<\/sub>2<\/sup> + y1<\/sub>2<\/sup> – 8x1<\/sub> + 6y1<\/sub> = 0
\nand QA2<\/sup> = x1<\/sub>2<\/sup> + y1<\/sub>2<\/sup> – 8x1<\/sub> + 6y1<\/sub> + 25
\n= 0 + 25 = 25
\n\u21d2 QA = 5
\n\u2234 Q (x1<\/sub>, y1<\/sub>) satisfy the geometric condition (1)
\n\u2234 Required equation of locus is x2<\/sup> + y2<\/sup> – 8x + 6y = 0
\nNote : Second part need not follow the problem from the definition of locus.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the equation of locus of a point which is equidistant from the points A (-3, 2) and B (0, 4). (V.S.A.Q)
\nAnswer:
\nLet P (x, y) be any point on the locus. Then from the given geometric condition PA = PB
\n\"TS
\n\u2234 PA2<\/sup> = PB2<\/sup>
\n\u21d2 (x + 3)2<\/sup> + (y – 2)2<\/sup> = (x – 0)2<\/sup> + (y – 4)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + 6x + 9 – 4y + 4 = x2<\/sup> + y2<\/sup> – 8y + 16
\n\u21d2 6x + 4y – 3 = 0
\n\u21d2 6x + 4y = 3 is the equation of the locus.<\/p>\n

Question 3.
\nFind the equation of locus of a point P such that the distance of P from the origin is twice the distance of P from A (1, 2).
\n(V.S.A.Q) (March 2012)
\nAnswer:
\nGiven O (0,0) and A (1,2) are the two points, P (x, y) be any point on the locus. From the given condition OP = 2PA
\n\"TS
\n\u21d2 OP2<\/sup> = 4 PA2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = 4[(x – 1)2<\/sup> + (y – 2)2<\/sup>]
\n\u21d2 x2<\/sup> + y2<\/sup> = 4 [(x2<\/sup> + y2<\/sup> – 2x – 4y + 1 + 4]
\n\u21d2 4 [x2<\/sup> + y2<\/sup> – 2x – 4y + 5]
\n\u21d2 4x2<\/sup> + 4y2<\/sup> – 8x – 16y + 20
\n\u2234 Equation to the locus of P is
\n3x2<\/sup> + 3y2<\/sup> – 8x – 16y + 20 = 0<\/p>\n

Question 4.
\nFind the equation of locus of a point which is equidistant from the coordinate axes. (V.S.A.Q)
\nAnswer:
\nLet P (x, y) be any point on the locus.
\nThe distance from P to X – axis is \u2019y and that of the distance to Y – axis is ‘x’.
\nGiven y = x \u21d2 y2<\/sup> = x2<\/sup>
\nlocus of P (x, y) is x2<\/sup> – y2<\/sup> = 0<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the equation of locus of a point equidistant from A (2, 0) and the Y – axis. (V.S.A.Q)
\nAnswer:
\nA (2, 0) is the given point and
\nLet P (x, y) be any point on the locus.
\nThe distance from P to Y – axis is PB = x
\nGiven PA = PB
\n\u21d2 PA2<\/sup> = PB2<\/sup>
\n\u21d2 (x – 2)2<\/sup> + y2<\/sup> = x2<\/sup>
\n\u21d2 x2<\/sup> – 4x + 4 + y2<\/sup> = x2<\/sup>
\n\u21d2 y2<\/sup> – 4x + 4 = 0
\n\u2234 Locus of P (x, y) is y – 4x + 4 = 0<\/p>\n

Question 6.
\nFind the equation of locus of a point P the square of whose distance from the origin is 4 times its y – coordinate. (V.S.A.Q)
\nAnswer:
\nLet P (x, y) be any point on the locus. Its distance from origin is OP
\nGiven that OP2<\/sup> = 4y
\n\u21d2 x2<\/sup> + y2<\/sup> = 4y
\n\u21d2 x2<\/sup> + y2<\/sup> – 4y = 0
\n\u2234 Equation to the locus of P is
\nx2<\/sup> + y2<\/sup> – 4y = 0<\/p>\n

Question 7.
\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| (V.S.A.Q)
\nAnswer:
\nLet P (x, y) be any point on the locus. Given A = (a, 0) and B = (-a, 0) are two points. Given condition is PA2<\/sup> + PB2<\/sup> = 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> = 2c2<\/sup> – 2a2<\/sup>
\n\u2234 x2<\/sup> + y2<\/sup> = c2<\/sup> – a2<\/sup> is the equation of locus of P.<\/p>\n

II.
\nQuestion 1.
\nFind the equation of locus of P, if the line segment joining (2, 3) and (-1,5) subtends a right angle at P. (May ’12, March ’13, ’05) (S.A.Q)
\nAnswer:
\n\"TS
\nA = (2, 3), B = (-1, 5) are the given points. P (x, y ) is any point on the locus.
\nGiven condition is \u2220APB = 90\u00b0
\nUsing pythagorous theorem, AP2<\/sup> + PB2<\/sup> = AB2<\/sup>
\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 x22<\/sup> – 4x + 4 + y2<\/sup> – 6y + 9 + x2<\/sup> + 2x + 1 + y2<\/sup> – 10y + 25 = 9 + 4
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 2x – 16y + 26 = 0
\n\u2234 Locus of P is x2<\/sup> + y2<\/sup> – x – 8y + 13 = 0
\n(x, y) \u2260 (2, 3) and (x, y) \u2260 (-1,5)<\/p>\n

\"TS<\/p>\n

Question 2.
\nThe ends of the hypotenuse of a right angled triangle are (0, 6 ) and ( 6, 0 ). Find the equation of locus of its third vertex. (S.A.Q)
\nAnswer:
\nThe ends of the hypotenuse are given as A (0, 6) and B (6, 0)
\nLet P (x, y) be the third vertex.
\n\"TS
\nGiven condition is \u2220APB = 90\u00b0
\n\u2234 By pythagorous theorem
\n\u21d2 AP2<\/sup> + PB2<\/sup> = AB2<\/sup>
\n\u21d2(x – 0)2<\/sup> + (y – 6)2<\/sup> + (x – 6)2<\/sup> + (y – 0)2<\/sup> = (0 – 6)2<\/sup> + (6 – 0)2<\/sup>
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 12y – 12x + 36 + 36 = 36 + 36
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 12x – 12y = 0
\n\u2234 Locus of P (x, y) is x2<\/sup> + y2<\/sup> – 6x – 6y = 0
\n(x, y) \u2260 (0, 6) and (x, y) \u2260 (6, 0)<\/p>\n

Question 3.
\nFind the equation of locus of a point the difference of whose distances from (- 5, 0) and (5, 0) is 8. (May 2011,’06, March 2001) (S.A.Q)
\nAnswer:
\nA (- 5, 0) and B (5, 0) are the given points.
\nLet P (x, y) be a point on the locus.
\nFrom the given condition |PA – PB| = 8 …… (1)
\nConsider
\nPA2<\/sup> – PB2<\/sup> = [(x + 5)2<\/sup> + (y – 0)2<\/sup>] – [(x – 5)2<\/sup> + (y – 0)2<\/sup>]
\n= (x2<\/sup> + 10x + 25 + y2<\/sup>) – (x2<\/sup> – 10x + 25 + y2<\/sup>) = – 20x
\n\u2234 (PA + PB) (PA – PB) = 20x
\n\u21d2 (PA + PB) (8) = 20x
\n\u21d2 PA + PB = \\(\\frac{5}{2}\\) x .
\nAdding (1) and (2)
\n2PA = 8 + \\(\\frac{5 x}{2}\\)x
\n\u21d2 4PA = 16 – 5x
\n\u21d2 16 PA = (16 – 5x)2<\/sup>
\n\u21d2 16 [(x + 5)2<\/sup> + y2<\/sup>] = (16 + 5x)2<\/sup>
\n\u21d2 16 [x2<\/sup> + y2<\/sup> + 10x + 25] = (16 + 5x)2<\/sup>
\n\u21d2 16x2<\/sup> + 16y2<\/sup> + 160x + 400 = 256 + 160x + 252<\/sup>
\n\u21d2 9x2<\/sup> + 16y2<\/sup> + 144 = 0
\n\u21d2 9x2<\/sup> – 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{9 x^2}{144}-\\frac{16 y^2}{144}\\) = 1
\n\u21d2 \\(\\frac{x^2}{16}-\\frac{y^2}{9}\\) = 1
\n\u2234 Equation of locus of P(x, y) is \\(\\frac{x^2}{16}-\\frac{y^2}{9}\\) = 1<\/p>\n

Question 4.
\nFind the equation of locus of P, if
\nA = (4, 0), B = (- 4, 0) and |PA – PB| = 4. (S.A.Q) (May’ 2007)
\nAnswer:
\nGiven that A = (4, 0) and B = (- 4, 0) are two points and let P (x ,y) be any point on the locus.
\nThe given condition is |PA – PB| = 4 ………………. (1)
\nConsider
\nPA2<\/sup> – PB2<\/sup> = [(x – 4)2<\/sup> + y2<\/sup>] – [(x + 4)2<\/sup> + y2<\/sup>]
\n= (x2<\/sup> – 8x + 16 + y2<\/sup>) – (x2<\/sup> + y2<\/sup> + 8x + 16)
\n= – 16x
\n(PA + PB) (PA – PB) = – 16x
\n\u21d2 (PA + PB) (4) = -16x
\n\u21d2 PA + PB = – 4x ……………….. (2)
\nAdding (1) and (2)
\n2PA = 4 – 4x
\n\u21d2 PA = 2 – 2x
\n\u21d2 PA2<\/sup> = (2 – 2x)2<\/sup>
\n\u21d2 (x – 4)2<\/sup> + y2<\/sup> = 4 – 8x + 4x2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> – 8x + 16 = 4x2<\/sup> – 8x + 4
\n\u21d2 3x2<\/sup> – y2<\/sup> = 12
\n\u21d2 \\(\\frac{x^2}{4}-\\frac{y^2}{12}\\) = 1
\n\u2234 Equation to the locus of P is \\(\\frac{x^2}{4}-\\frac{y^2}{12}\\) = 1<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the equation of locus of a point, the sum of whose distances from (0, 2) and (0, – 2) is 6 . (S.A.Q)
\nAnswer:
\nLet A = (0, 2) and B = (0, – 2) are the two given points.
\nLet P (x, y) b.e any point on the locus.
\nFrom the given condition, PA + PB = 6 …………….. (1)
\nConsider
\nPA2<\/sup> – PB2<\/sup> = [(x – 0)2<\/sup> + (y – 2)2<\/sup>] – [(x – 0)2<\/sup> + (y + 2)2<\/sup>]
\n= x2<\/sup> + y2<\/sup> – 4y + 4 – x2<\/sup> – y2<\/sup> – 4y – 4 = – 8y
\n\u2234 (PA + PB) (PA – PB) = – 8y
\n\u21d2 6 (PA – PB) = – 8y
\n\u21d2 (PA – PB) = \\(\\frac{-8 y}{6}=\\frac{-4 y}{3}\\) ………………… (2)
\nAdding (1) and (2)
\n\"TS
\n\u21d2 9x2<\/sup> + 9y2<\/sup> + 36 = 81 + 4y2<\/sup>
\n\u21d2 9x2<\/sup> + 5y2<\/sup> = 45
\n\u21d2 \\(\\frac{x^2}{5}+\\frac{y^2}{9}\\) = 1
\n\u2234 Equation to the locus of P is \\(\\frac{x^2}{5}+\\frac{y^2}{9}\\) = 1<\/p>\n

Question 6.
\nFind the equation of the locus of P, if A = ( 2, 3 ), B = ( 2, -3 ) and PA + PB = 8. (SA.Q)
\nAnswer:
\nLet P (x, y ) be any point on the locus.
\nGiven condition is PA + PB = 8 (1)
\nPA2<\/sup> – PB2<\/sup> = [(x – 2)2<\/sup> + (y – 3)2<\/sup>] – [ (x – 2)2<\/sup> + ( y + 3)2<\/sup>]
\n= (x – 2)2<\/sup> + (y – 3)2<\/sup> – (x – 2)2<\/sup> – (y + 3)2<\/sup> = – 12y
\n(y – 3)2<\/sup> – (y + 3)2<\/sup> = – 12y
\n\u2234 PA2<\/sup> – PB2<\/sup> = – 12y
\n(PA + PB) (PA – PB) = -12y
\n\u21d2 8 (PA – PB) = – 12y
\n\u21d2 PA – PB = \\(\\frac{-12 y}{8}=\\frac{-3 y}{2}\\) ……………….. (2)
\nAdding (1) and (2)
\n2PA = 8 – \\(\\frac{3 \\mathrm{y}}{2}\\) = \\(\\frac{16-3 y}{2}\\)
\n\u21d2 4PA = 16 – 3y
\n\u21d2 16PA2<\/sup> = (16 – 3y)2<\/sup>
\n\u21d2 16 [(x – 2)2<\/sup> + (y – 3)2<\/sup>] = (16 – 3y)2<\/sup>
\n\u21d2 16 [x2<\/sup> + y2<\/sup> – 4x – 6y + 13] = 256 – 96y + 9y2<\/sup>
\n\u21d2 16x2<\/sup> + 7y2<\/sup> – 64x – 48 = 0
\n\u2234 Equation to the locus of P is
\n16x2<\/sup> + 7y2<\/sup> – 64x – 48 = 0
\n(or) 16(x2<\/sup> – 4x) + 7y2<\/sup> = 48
\n\u21d2 16 (x2<\/sup> – 4x + 4) + 7y2<\/sup> = 112
\n\u21d2 16 (x – 2)2<\/sup> + 7y2<\/sup> = 112.
\n\u21d2 \\(\\frac{(x-2)^2}{7}+\\frac{y^2}{16}\\)<\/p>\n

\"TS<\/p>\n

Question 7.
\nA (5, 3) and B (3, – 2) are two fixed points. Find the equation to the locus of P, so that the area of triangle PAB is 9. (S.A.Q) (March 2006)
\nAnswer:
\nA (5, 3 ), B (3, -2 ) are the given points.
\nLet P (x, y ) be any point on the locus. Given condition is that the area of \u2206 PAB = 9
\n\u21d2 \\(\\frac{1}{2}\\) |[x1<\/sub> (y2<\/sub> – y3<\/sub>) + x2<\/sub> (y3<\/sub> – y1<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)[|
\n= \\(\\frac{1}{2}\\) |[5(- 2 – y) + 3 (y – 3) + x (3 + 2)]| = 9
\n\u21d2 5x – 2y – 19 = \u00b118
\n\u21d2 5x – 2y – 19 = 18 (or) 5x – 2y – 19 = – 18
\n\u21d2 5x – 2y – 37 = 0 (or) 5x – 2y – 1= 0
\n\u2234 Locus of P is (5x – 2y – 37) (5x – 2y – 1) = 0<\/p>\n

Question 8.
\nFind the equation of locus of a point, which forms a triangle of area 2 with the points A (1, 1 ) and B ( – 2, 3 ). (S.A.Q)
\nAnswer:
\nA ( 1, 1), B(- 2, 3 ) are the two given points and Let P ( x, y ) be any point on the locus.
\nGiven condition is \u2206 PAB = 2
\n\u21d2 \u2234 \\(\\frac{1}{2}\\) |1(3 – y) – 2(y – 1) + x(1 – 3)| = 2
\n\u21d2 |3 – y – 2y + 2 – 2x| = 4
\n\u21d2 – 2x – 3y + 5 = \u00b14
\n\u21d2 – 2x – 3y + 5 = 4 (or) – 2x – 3y + 5 = – 4
\n\u21d2 2x + 3y – 1 = 0 (or) 2x + 3y – 9 = 0
\n\u2234 Locus of P (x, y) is (2x + 3y – 1) (2x + 3y – 9) = 0<\/p>\n

Question 9.
\nIf the distance from P to the points (2, 3) and ( 2, – 3 ) are in the ratio 2 : 3 then find the equation of locus of P. (S.A.Q) (May 2014, March 2014)
\nAnswer:
\nLet P (x, y ) be any point on the locus.
\nGiven points are A (2, 3) and B (2, -3) and given condition is PA : PB = 2: 3
\n\u21d2 3PA = 2PB \u21d2 9PA2<\/sup> = 4PB2<\/sup>
\n\u21d2 9 [(x – 2)2<\/sup> + (y – 3)2<\/sup>] = 4 [(x – 2)2<\/sup> + (y + 3)2<\/sup>]
\n\u21d2 9 [(x2<\/sup> – 4x + 4 + y2<\/sup> – 6y + 9)] = 4 [x2<\/sup> – 4x + 4 + y2<\/sup> + 6y + 9]
\n\u21d2 5x2<\/sup> + 5y2<\/sup> – 20x – 78y + 65 = 0
\n\u2234 Equation to the locus of P is
\n5x2<\/sup> + 5y2<\/sup> – 20x – 78y + 65 = 0<\/p>\n

Question 10.
\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 (S.A.Q) (May 2007)
\nAnswer:
\nLet P (x, y) be any point on the locus. Given points are
\nA = (1, 2) ; B = (2, – 3) and C = (- 2, 3)
\nGiven 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>]
\n= 2 [(x + 2)2<\/sup> + (y – 3)2<\/sup>]
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 6x + 2y + 18
\n= 2x2<\/sup> + 2y2<\/sup> + 8x – 12y + 26
\n\u21d2 14x – 14y + 8 = 0
\n\u21d2 7x – 7y + 4 = 0
\n\u2234 Equation to the locus of P is
\n7x – 7y + 4 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these TS Intermediate Maths 1B Solutions Chapter 1 Locus Ex 1(a) to find a better approach to solving the problems. TS Inter 1st Year Maths 1B Locus Solutions Exercise 1(a) Question 1. Find the equation of locus of a point which is at a distance 5 from A (4, – 3). (V.S.A.Q) … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/3808"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=3808"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/3808\/revisions"}],"predecessor-version":[{"id":5945,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/3808\/revisions\/5945"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=3808"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=3808"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=3808"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}