{"id":20217,"date":"2020-01-18T17:53:06","date_gmt":"2020-01-18T12:23:06","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20217"},"modified":"2021-07-02T15:28:47","modified_gmt":"2021-07-02T09:58:47","slug":"2nd-puc-basic-maths-question-bank-chapter-16-ex-16-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-16-ex-16-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 16 Parabola Ex 16.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 16.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 16 Parabola Ex 16.1<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Parabola Ex 16.1 One or Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nWrite the characteristics of the following parabolas.
\n(a) y2<\/sup> = 16x
\n(b) y2<\/sup> = -8x
\n(c) 3x2<\/sup> = -8y
\n(d) x2<\/sup> = 8y
\n(e) y2<\/sup> = 8x
\n(f) y2<\/sup> + 4x = 0
\n(g) 3x2<\/sup> + 4y = 0
\n(h) x2<\/sup> + 4x = 0
\nAnswger:
\n(a) Compare y2<\/sup> = 16x with y2<\/sup> = 4ax
\n\"2nd
\nWe get 4a = 16 \u21d2 a = 4
\nThe curve turns right side with vertex (0,0)
\nFocus S(a,0) = (4,0)
\nDirectrix x = -a or x = -4 or x + 4 = 0
\nAxis x-axis (equation is y = 0)
\nTangent y-axis (equation is x = 0)
\nEquation of LR is x = a \u21d2 x = 4 or x – 4 = 0
\nLength of LR 4a = 16
\nEnds of LR (a, 2a) (a, -2a) = (4,8) and (4,-8)<\/p>\n

\"KSEEB<\/p>\n

(b) y2<\/sup> = -8x compare with y2<\/sup> = -4ax
\n\"2nd
\nThe curve turns left hand side and 4a = 8 \u21d2 a = 2
\nVertex = (0,0)
\nAxis x-axis (equation is y = 0)
\nTangent y-axis (equation is x = 0)
\nFocus S(-2,0)
\nDirectrix, x = 2 or x – 2 = 0
\nEquation of LR x = -2 or x + 2 = 0
\nLength of LR = 4a = 8
\nEnds of LR (-a, 2a) (-a, -2a) = (-2, 4) (-2,-4)<\/p>\n

(C) 3x2<\/sup> = -8y \u21d2 x2<\/sup> = \\(\\frac {-8 }{ 3 }\\) y compare this with
\n\"2nd
\nx2<\/sup> = -4ay \u21d2 4a = \\(\\frac { 8 }{ 3 }\\) \u21d2 a = \\(\\frac { 2 }{ 3 }\\)
\nThe curve turns downwards
\nVertex V = (0,0)
\nAxis y-axis (equation is x = 0)
\nFocus S = \\(\\left(0,-\\frac{2}{3}\\right)\\)
\nTangent x-axis (equation is y = 0)
\nDirectrix y = \\(\\frac { 2 }{ 3 }\\) = or 3y – 2 = 0
\nEquation of LR y = –\\(\\frac { 2 }{ 3 }\\) or 3y + 2 = 0
\nLength of LR = 42 = \\(\\frac { 8 }{ 3 }\\) ; Ends of LR \\(\\left(\\frac{4}{3},-\\frac{2}{3}\\right)\\left(-\\frac{4}{3},-\\frac{2}{3}\\right)\\)<\/p>\n

\"KSEEB<\/p>\n

(d) x2<\/sup> = 8y compare this with x2<\/sup> = 4ay the curve turns upwards
\n\"2nd
\n4a = 8 \u21d2 a = 2
\nVertex, V = (0,0)
\nAxis y-axis (equation is x = 0)
\nTangent x-axis (equation is y = 0)
\nDirectrix y = -2 or y + 2 = 0
\nEquation of LR y = 2 or y – 2 = 0
\nLength of LR 4a = 8
\nEnds of LR (4, 2) (-4, 2)<\/p>\n

(e) y2<\/sup> = 8x compare with y2<\/sup> = 4ax. The curve turns right side
\n\"2nd
\n4a = 8 \u21d2 a = 2
\nVertex, V = (0,0)
\nAxis, x-axis (equation is y = 0)
\nTangent, y-axis (equation is x = 0)
\nFocus S = (2,0)
\nDirectrix x = -2 or x + 2 = 0
\nEquation of LR x = 2
\nLength of LR 4a = 8
\nEnds of LR (2, 4) (2,-4)<\/p>\n

\"KSEEB<\/p>\n

(f) Compare y2<\/sup> = -4x with y2<\/sup> = -4ax. The curve turns
\n\"2nd
\nLeft hand side
\n4a = 4 \u21d2 a = 1
\nVertex, V = (0,0)
\nAxis x-axis (equation is y = 0)
\nTangent y-axis (equation is x = 0)
\nDirectrix, x = 1 or x = -1 = 0
\nEquation of LR x = -1 or x + 1 = 0
\nEnds of LR (-1,2) (-1, -2)
\nLength of LR = 4a = 4<\/p>\n

(g) 3x2<\/sup> + 4y = 0
\n\"2nd
\nx2<\/sup> = \\(\\frac{-4}{3} y\\) compare with x2<\/sup> = – 4ay.
\nThe curve turns downwards and 4a = \\(\\frac { 4 }{ 3 }\\) \u21d2 a = \\(\\frac { 1 }{ 3 }\\)
\nVertex, V = (0,0)
\nAxis y-axis (equation is x = 0)
\nTangent x-axis (equation is y = 0)
\nFocus, s=(0,- \\(\\frac { 1 }{ 3 }\\)) Directrix y = \\(\\frac { 1 }{ 3 }\\) or 3y – 1=0
\nEquation of LR y = \\(\\frac { -1 }{ 3 }\\) or 3y +1=0 Length of LR = 4a = \\(\\frac { 4 }{ 3 }\\)
\nEnds of LR \\(\\left(-\\frac{2}{3},-\\frac{1}{3}\\right)\\left(\\frac{2}{3},-\\frac{1}{3}\\right)\\)<\/p>\n

\"KSEEB<\/p>\n

(h) x2<\/sup> + 16y = 0 compare with x2<\/sup> = -4ay
\n\"2nd
\nx2<\/sup> = -16y. The curve turns downward
\n4a = 16 \u21d2 a = 4
\nVertex, V = (0,0)
\nAxis y-axis (Equation is x = 0)
\nTangent x-axis (Equation is y = 0)
\nFocus, S = (0, -4)
\nDirectrix y = 4 or y – 4 = 0
\nEquation of LR y = -4 or y + 4 = 0
\nLength of LR = 4a = 16
\nEnds of LR (-8,-4) (8,-4).<\/p>\n

Question 2.
\nIf the length of the latus rectum of y2<\/sup> = 8kx is 4 find k
\nAnswer:
\nCompare y2<\/sup> = 8kx with y2<\/sup> = 4ax
\nWe get 4a = 8k \u21d2 a = 2k
\nGiven length of LR = 4
\n4a = 4 \u21d2 8k = 4 \u21d2 k = \\(\\frac { 1 }{ 2 }\\)<\/p>\n

Question 3.
\nIf the length of the latus rectum of x2<\/sup> = 4ky is 8, find the value of k.
\nAnswer:
\nCompare x2<\/sup> = 4ky with x2<\/sup> = 4ay.
\nWe get 4k = 4a \u21d2 a=k
\nGiven length of LR = 8
\n4a = 8
\n4k = 8
\nk = 2.<\/p>\n

\"KSEEB<\/p>\n

Part-B<\/span><\/p>\n

2nd PUC Basic Maths Parabola Ex 16.1 Two or Three Marks Question and Answers<\/strong><\/p>\n

Question 1.
\nFind the equation of the parabola given that its.
\n(a) Vertex is (0,0) and focus is (4,0)
\n(b) Vertex is (0,0) and directrix is y = – 3
\n(c) Focus is (1,0) and directrix is x = – 1,
\n(d) Focus is (-4, 0) and directrix is x = 4
\n(e) Focus is (0, -3) and directrix is y=3
\n(f) Focus is (0, 6) and vertex is (0,0)
\n(g) Vertex is (0,0), axis is y axis and passes through \\(\\left(\\frac{1}{2}, 2\\right)\\)
\n(h) Vertex is (0,0) axis is y axis and passes through (-1, -3)
\nAnswer:
\n(a) Given V = (0,0) and Focus, S = (4,0)
\nIt is a right handed parabola with
\na = 4 and standard form is
\ny2<\/sup> = 4ax
\nPut a = 4, then
\ny2<\/sup> = 16x.
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(b) Given V = (0,0) and directrix = y = -3
\nFocus = (0,3) \u21d2 a = 3
\nIt is a upward parabola and
\nstandard form is x2<\/sup> = 4ay
\nPut a = 3, x2<\/sup> = 4.3y
\n\u21d2 x2<\/sup> = 12y
\n\"2nd<\/p>\n

(C) Given Focus = S = (1,0) and directrix is
\nx =-15 \u21d2 a = 1
\nit is a right handed parabola
\nand standard form is y2<\/sup> = 4ax put a = 1
\n\u21d2 y2<\/sup> = 4x
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(d) Give Focus = S = (-4, 0) and directrix
\nis x = 4; S = (-4,0) \u21d2 a = 4
\nIt is a left handed parabola standard
\nform is y2<\/sup> = -4ax put a = 4
\nwe get y2<\/sup> = -16x
\n\"2nd<\/p>\n

(e) Given focus is (0, -3) and directrix is y = 3
\n\u21d2 a = 3 and curve turns
\ndownwards and standard form is
\nx2<\/sup> = -4ay put a = 3,
\nwe get x2<\/sup> = -12y
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(f) Given focus S = (0,6) and Vertex, V = (0,0)
\n\u21d2 a = 6 and it is a upward parabola and
\nthe standard form is x2<\/sup> = 4ay
\nPut a = 6
\nWe get x2<\/sup> = 24y
\n\"2nd<\/p>\n

(g) Given vertex V = (0,0) and axis is y-axis and it passes through \\(\\left(\\frac{1}{2}, 2\\right)\\)
\nwhich is in I quadrant
\n\u2234 it is a upward parabola
\nx2<\/sup> = 4ay put x = \\(\\frac { 1 }{ 2 }\\) and y = 2; \\(\\frac { 1 }{ 4 }\\) = 4a 2 \u21d2 \\(\\frac { 1 }{ 4 }\\) = 8a \u21d2 a = \\(\\frac { 1 }{ 32 }\\)
\n\u2234 The required equation is x2<\/sup> = 4ay
\nx2<\/sup> =4. \\(\\frac{1}{32} y\\) \u21d2 x2<\/sup> = \\(\\frac{1}{8} y\\) or 8x2<\/sup> – y = 0<\/p>\n

(h) Vertex, V(0, 0) and y-axis and it passes through the point (-1, -3)
\nwhich is in third quadrant
\n\u2234 It is a downward parabola,
\ni.e., x2<\/sup> = -4ay
\nput x=-1 and y = -3; (-1)2<\/sup> = -4a(-3)
\n1 = 12a \u21d2 a = \\(\\frac { 1 }{ 12 }\\)
\n\u2234 Required equation is x2<\/sup> = -4ay
\nx2<\/sup> = -4.\\(\\frac { 1 }{ 12 }y\\) ; 3x2<\/sup> + y = 0.<\/p>\n

\"KSEEB<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Parabola Ex 16.1 Five Mark Questions and Answers<\/strong><\/p>\n

Question 1.
\nDefine parabola and with usual notation Prove that y2<\/sup> = 4ax geometrically
\nSol.
\nDefinition of Parabola and other forms of parabola
\nParabola is a conic section whose eccentricity is 1. That is, Parabola is the locus of the point which moves such that its distance from the fixed point (called the focus) is equal to its distance from the fixed line (called the directrix) We shall derive the equation of the parabola in its standard form.
\n\"2nd<\/p>\n

Theorem: The equation of a Parabola with proper choice of co-ordinates axes is y2<\/sup> = 4ax
\nProof: Let S be the focus and the line l be the directrix. Draw SZ through S and perpendicular to the directrix. Let O be the midpoint of the line segment SZ.<\/p>\n

Let SZ = 2a (a > 0) \u21d2 OZ = OS = a
\nWe shall choose 0 as the origin and the line ZOS as x axis.
\nThe line YOY’ through O, perpendicular to the x-axis will be y-axis.
\nWith this choice of co-ordinate axes we have S(a, o) and Z(-a, 0)
\nThe equation of the directrix is l is x = -a
\nLet P(x,y) be any point on the Parabola. Then by the definition of the Distance of P
\nfrom S = Distance of P from the line ‘l’.
\n\"2nd
\n|SP| = |PM|
\nSP2<\/sup> = ZN2<\/sup> since PM = ZN
\nSP2<\/sup> = (OZ + ON)2<\/sup>
\n(x – a)2<\/sup> + y2<\/sup> = (a + x)2<\/sup> (|OZ| = a, |ON| = x)
\nx2<\/sup> – 2ax + a2<\/sup> + y2<\/sup> = a2<\/sup> + 2ax + x2<\/sup>
\ny2<\/sup> = 4ax which is the equation of the parabola.<\/p>\n

\"KSEEB<\/p>\n

Note:
\nThe equation y2<\/sup> = 4ax is also called standard form of the equation of the parabola.
\nShape of the Parabola y2<\/sup> = 4ax
\nWe shall note few observations from the equation y2<\/sup> = 4ax, which will help us to trace the curve parabola.
\n1. Ify is replaced by -y in the equation remains same, i.e., (-y)2<\/sup> = 4ax \u2192 y2<\/sup> = 4ax. This shows that if (x, y) is any point on the curve y2<\/sup> = 4ax, then (x, -y) is also a point on the curve. Thus, the curve is symmetric about the x – axis, i.e., the shape of the curve above the x axis is the mirror image (about the x axis) of the shape of the curve below the x – axis.<\/p>\n

2. If x < 0, then y will be a negative quantity (note that a > 0) and therefore y2<\/sup> = 4ax will have no real solution for y. This shows that no part of the curve lies to the left side of the x-axis.<\/p>\n

3. If y =0, the only value of x we get is zero. Thus the curve cuts the x axis at the origin(0,0)<\/p>\n

4. If x = 0, we get y2<\/sup> = 0, which gives y = 0. Thus the curve cuts the y-axis at the origin and further the y-axis meets the curve only at the origin. That is, y – axis is a tangent to the curve at the origin.<\/p>\n

5. For any point P(x, y) on the parabola we have
\ny2<\/sup> = 4ax \u21d2 y = \u00b12\\(\\sqrt{\\mathrm{ax}}\\)
\ny = 2\\(\\sqrt{\\mathrm{ax}}\\) and y = -2\\(\\sqrt{\\mathrm{ax}}\\)
\nThis shows that, as x increases from 0 to \u221e, y also increases from 0 to \u221e or y decreases from 0 to -\u221e. Thus, the two branches of the parabola, laying on opposite sides of the x – axis, will extend to infinity towards the positive directions of the x – axis.<\/p>\n

From the above discussion and by plotting few points, whose coordinates satisfy y2<\/sup> = 4ax, it is found the shape of the parabola is as shown in the following figure.
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

The origin 0 is called the Vertex of the parabola, y2<\/sup> = 4ax, it is also denoted by V(0,0).
\nThe line ZSX is called the axis of the parabola and its equation is y = 0
\nThe focus S (a,0) and the equation of the directrix is x – a i.e., X + a = 0 They – axis is called the tangent at the vertex – its equation is x = 0.<\/p>\n

Note:<\/p>\n

The distance between the vertex and the focus is the distance between the vertex and the directrix is equal to a and the distance between the directrix and the focus is 2 a.<\/p>\n

Definition:
\nThe chord passing through the focus and perpendicular to the axis of the parabola is called the latus rectum of the parabola.<\/p>\n

In the figure LSL1<\/sup> is the latus rectum.
\nThe points L and L1<\/sup> on the parabola are called end points of the latus rectum, the length is called the length of the latus rectum.
\nClearly, the x – coordinates of L and L’ is a because OS = a. To find the corresponding coordinates, we shall put x = a in y2<\/sup> = 4ax, we get y = \u00b12a. Thus, y – coordinates of L and L1<\/sup> respectively.
\nL = (a, 2a) and L1<\/sup> = (a, -2a)
\n\u2234 consider |LL’| = \\(\\sqrt{(a-a)^{2}+(2 a+2 a)^{2}}=\\sqrt{(4 a)^{2}}=4 a\\)
\nThus, the length of the latus rectum, LL1<\/sup> = 4a.<\/p>\n

Note :
\nAny chord passing through the focus is called a focal chord. Focal chord need not be perpendicular to the axis of the parabola<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 16.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 16 Parabola Ex 16.1 Part – A 2nd PUC Basic …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20217"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=20217"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20217\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20217"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20217"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20217"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}