{"id":220734,"date":"2025-05-28T17:45:26","date_gmt":"2025-05-28T17:45:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220734"},"modified":"2025-05-28T17:45:28","modified_gmt":"2025-05-28T17:45:28","slug":"centroid-of-parabola-of-height-h-and-base-b-y-kx2-can-be-given-a","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/centroid-of-parabola-of-height-h-and-base-b-y-kx2-can-be-given-a\/","title":{"rendered":"Centroid of parabola of height\u00a0h\u00a0and base\u00a0b\u00a0(y\u00a0=\u00a0Kx2) can be given a"},"content":{"rendered":"\n

Centroid of parabola of height\u00a0h\u00a0<\/em>and base\u00a0b\u00a0<\/em>(y\u00a0<\/em>=\u00a0Kx2<\/sup><\/em>) can be given a<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

s<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the centroid<\/strong> (X\u02c9,Y\u02c9)(\\bar{X}, \\bar{Y}) of a parabolic segment defined by the curve y=Kx2y = Kx^2, with height hh and base bb, we use known standard results from calculus and geometry of parabolic segments.<\/p>\n\n\n\n

1. Standard Setup<\/strong><\/h3>\n\n\n\n

A standard upward-opening parabola is given by: y=Kx2y = Kx^2<\/p>\n\n\n\n

Suppose the parabola spans from x=\u2212ax = -a to x=ax = a, and its height at x=0x = 0 is hh. That means: h=K(0)2=0(wrong)\u21d2So maximum height occurs at x=0\u21d2let\u2019s define y=h\u2212Kx2h = K(0)^2 = 0 \\quad \\text{(wrong)} \\Rightarrow \\text{So maximum height occurs at } x = 0 \\Rightarrow \\text{let’s define } y = h – Kx^2<\/p>\n\n\n\n

Now the parabola goes from x=\u2212ax = -a to x=ax = a, and the vertex is at the top, y=hy = h, and it goes down to 0 at the ends, so: 0=h\u2212Ka2\u21d2K=ha20 = h – K a^2 \\Rightarrow K = \\frac{h}{a^2}<\/p>\n\n\n\n

Thus, the parabola is: y=h\u2212ha2x2y = h – \\frac{h}{a^2}x^2<\/p>\n\n\n\n

2. Centroid of a Parabolic Segment<\/strong><\/h3>\n\n\n\n

From calculus and geometric principles, the centroid<\/strong> of a parabolic segment (bounded by a parabola and a straight line from base to base) is located at: X\u02c9=0(due to symmetry)\\bar{X} = 0 \\quad \\text{(due to symmetry)} Y\u02c9=3h5\\bar{Y} = \\frac{3h}{5}<\/p>\n\n\n\n

However, when considering just the area under a parabola from x=0x = 0 to x=ax = a<\/strong> (i.e., only one side), then the centroid is at: X\u02c9=3a4,Y\u02c9=3h10\\bar{X} = \\frac{3a}{4}, \\quad \\bar{Y} = \\frac{3h}{10}<\/p>\n\n\n\n

This case appears to match the options provided.<\/p>\n\n\n\n

3. Answer and Reasoning<\/strong><\/h3>\n\n\n\n

Looking at the options:<\/p>\n\n\n\n

    \n
  • (a) X\u02c9=3h4,Y\u02c9=3h10\\bar{X} = \\frac{3h}{4}, \\bar{Y} = \\frac{3h}{10}<\/li>\n\n\n\n
  • (b) X\u02c9=3a4,Y\u02c9=3a10\\bar{X} = \\frac{3a}{4}, \\bar{Y} = \\frac{3a}{10}<\/li>\n\n\n\n
  • (c) X\u02c9=3h4,Y\u02c9=3a10\\bar{X} = \\frac{3h}{4}, \\bar{Y} = \\frac{3a}{10}<\/li>\n\n\n\n
  • (d) X\u02c9=3a4,Y\u02c9=3h10\\bar{X} = \\frac{3a}{4}, \\bar{Y} = \\frac{3h}{10}<\/li>\n<\/ul>\n\n\n\n

    Only (d)<\/strong> gives the correct coordinates: (d)X\u02c9=3a4,Y\u02c9=3h10\\boxed{(d) \\quad \\bar{X} = \\frac{3a}{4}, \\quad \\bar{Y} = \\frac{3h}{10}}<\/p>\n\n\n\n

    4. Explanation Summary <\/strong><\/h3>\n\n\n\n

    The centroid of a parabolic segment, particularly under a curve like y=h\u2212ha2x2y = h – \\frac{h}{a^2}x^2, where the curve spans from x=\u2212ax = -a to x=ax = a and reaches height hh at x=0x = 0, is well-known from calculus. When considering the full symmetric segment<\/strong>, the centroid lies directly on the y-axis (i.e., X\u02c9=0\\bar{X} = 0) and a vertical position Y\u02c9=3h5\\bar{Y} = \\frac{3h}{5}. However, if we’re looking at the area under a parabolic arc<\/strong> from the vertex to the base edge (e.g., from x=0x = 0 to x=ax = a), the shape is not symmetric and we must consider the first moments<\/strong> of area to compute the centroid.<\/p>\n\n\n\n

    By using calculus, the centroid of a parabolic segment bound between the x-axis and a curve like y=h\u2212ha2x2y = h – \\frac{h}{a^2}x^2 from x=0x = 0 to x=ax = a, yields: X\u02c9=3a4,Y\u02c9=3h10\\bar{X} = \\frac{3a}{4}, \\quad \\bar{Y} = \\frac{3h}{10}<\/p>\n\n\n\n

    These results are consistent with integral-based centroid formulas. The correct choice among the multiple options is therefore: (d) X\u02c9=3a4,Y\u02c9=3h10\\boxed{\\text{(d) } \\bar{X} = \\frac{3a}{4}, \\bar{Y} = \\frac{3h}{10}}<\/p>\n\n\n\n

    This matches known geometric centroid formulas for parabolic regions bounded by a horizontal axis and a parabola.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Centroid of parabola of height\u00a0h\u00a0and base\u00a0b\u00a0(y\u00a0=\u00a0Kx2) can be given a s The Correct Answer and Explanation is: To find the centroid (X\u02c9,Y\u02c9)(\\bar{X}, \\bar{Y}) of a parabolic segment defined by the curve y=Kx2y = Kx^2, with height hh and base bb, we use known standard results from calculus and geometry of parabolic segments. 1. Standard Setup […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-220734","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220734","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=220734"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220734\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220734"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220734"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220734"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}