{"id":208442,"date":"2025-04-26T22:27:43","date_gmt":"2025-04-26T22:27:43","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=208442"},"modified":"2025-04-26T22:27:45","modified_gmt":"2025-04-26T22:27:45","slug":"a-latus-rectum-of-an-ellipse-is-a-chord-passing-through-the-foci-and-perpendicular-to-the-major-axis","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/26\/a-latus-rectum-of-an-ellipse-is-a-chord-passing-through-the-foci-and-perpendicular-to-the-major-axis\/","title":{"rendered":"A latus rectum of an ellipse is a chord passing through the foci and perpendicular to the major axis"},"content":{"rendered":"\n

A latus rectum of an ellipse is a chord passing through the foci and perpendicular to the major axis. Suppose that an ellipse has major and minor axes of lengths 2a and 2b, where a>b.<\/p>\n\n\n\n

a) (3 marks) Show that the length of a latus rectum of an ellipse is 2b2\/a. Hint: The foci of an ellipse lie on the major axis,units from the centre of the ellipse, witha22 having equation Give all answers in exact form<\/p>\n\n\n\n

(b) (3 marks) Find the endpoints of the latera recta (plural of latus rectum) for the ellipse 13r2 + 26x +4y2-12y -30.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Alright, let’s break it down carefully:<\/p>\n\n\n\n

\n\n\n\n

Part (a): Show that the length of the latus rectum is ( \\frac{2b^2}{a} )<\/strong><\/h3>\n\n\n\n

The standard form<\/strong> of an ellipse centered at the origin with a horizontal major axis is:<\/p>\n\n\n\n

[
\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1
\\quad \\text{where} \\quad a > b > 0
]<\/p>\n\n\n\n

The foci<\/strong> are located at ( (c, 0) ) and ( (-c, 0) ), where:<\/p>\n\n\n\n

[
c = \\sqrt{a^2 – b^2}
]<\/p>\n\n\n\n

The latus rectum<\/strong> is a vertical line segment passing through a focus and perpendicular to the major axis.<\/p>\n\n\n\n

Suppose we focus on the focus at ( (c, 0) ).
Since the latus rectum is vertical<\/strong> at ( x = c ), substitute ( x = c ) into the ellipse equation:<\/p>\n\n\n\n

[
\\frac{c^2}{a^2} + \\frac{y^2}{b^2} = 1
]<\/p>\n\n\n\n

We know ( c^2 = a^2 – b^2 ), so:<\/p>\n\n\n\n

[
\\frac{a^2 – b^2}{a^2} + \\frac{y^2}{b^2} = 1
]
[
1 – \\frac{b^2}{a^2} + \\frac{y^2}{b^2} = 1
]
[
\\frac{y^2}{b^2} = \\frac{b^2}{a^2}
]
[
y^2 = \\frac{b^4}{a^2}
]
[
y = \\pm \\frac{b^2}{a}
]<\/p>\n\n\n\n

Thus, the endpoints<\/strong> of the latus rectum at focus ( (c, 0) ) are:<\/p>\n\n\n\n

[
\\left(c, \\frac{b^2}{a}\\right) \\quad \\text{and} \\quad \\left(c, -\\frac{b^2}{a}\\right)
]<\/p>\n\n\n\n

The length<\/strong> of the latus rectum is the vertical distance between these two points:<\/p>\n\n\n\n

[
\\text{Length} = \\left( \\frac{b^2}{a} – \\left(-\\frac{b^2}{a}\\right) \\right) = \\frac{2b^2}{a}
]<\/p>\n\n\n\n

\u2705 Hence, proved<\/strong> that the length of a latus rectum is ( \\boxed{\\frac{2b^2}{a}} ).<\/p>\n\n\n\n

\n\n\n\n

Part (b): Find the endpoints of the latera recta for the ellipse ( 13x^2 + 26x + 4y^2 – 12y – 30 = 0 )<\/strong><\/h3>\n\n\n\n

First, rewrite the given ellipse in standard form<\/strong>:<\/p>\n\n\n\n

Group ( x )-terms and ( y )-terms:<\/p>\n\n\n\n

[
13(x^2 + 2x) + 4(y^2 – 3y) = 30
]<\/p>\n\n\n\n

Complete the square:<\/p>\n\n\n\n

    \n
  • For ( x )-terms: ( x^2 + 2x \\to (x+1)^2 – 1 )<\/li>\n\n\n\n
  • For ( y )-terms: ( y^2 – 3y \\to (y – \\frac{3}{2})^2 – \\frac{9}{4} )<\/li>\n<\/ul>\n\n\n\n

    Substituting:<\/p>\n\n\n\n

    [
    13\\left((x+1)^2 – 1\\right) + 4\\left(\\left(y – \\frac{3}{2}\\right)^2 – \\frac{9}{4}\\right) = 30
    ]
    Expand:<\/p>\n\n\n\n

    [
    13(x+1)^2 – 13 + 4\\left(y-\\frac{3}{2}\\right)^2 – 9 = 30
    ]
    [
    13(x+1)^2 + 4\\left(y-\\frac{3}{2}\\right)^2 – 22 = 30
    ]
    [
    13(x+1)^2 + 4\\left(y-\\frac{3}{2}\\right)^2 = 52
    ]
    Divide by 52:<\/p>\n\n\n\n

    [
    \\frac{(x+1)^2}{4} + \\frac{(y – \\frac{3}{2})^2}{13} = 1
    ]<\/p>\n\n\n\n

    Now it’s in standard form:
    [
    \\frac{(x+1)^2}{4} + \\frac{(y-\\frac{3}{2})^2}{13} = 1
    ]<\/p>\n\n\n\n

    Compare with:<\/h4>\n\n\n\n

    [
    \\frac{(x-h)^2}{a^2} + \\frac{(y-k)^2}{b^2} = 1
    ]<\/p>\n\n\n\n

    Thus:<\/p>\n\n\n\n

      \n
    • Center ( (h, k) = (-1, \\frac{3}{2}) )<\/li>\n\n\n\n
    • ( a^2 = 4, \\quad b^2 = 13 )<\/li>\n<\/ul>\n\n\n\n

      Notice b > a<\/strong>, so the major axis is vertical<\/strong>!<\/p>\n\n\n\n

      \n\n\n\n

      Step 1: Find ( c ) (distance to foci):<\/h4>\n\n\n\n

      [
      c = \\sqrt{b^2 – a^2} = \\sqrt{13 – 4} = \\sqrt{9} = 3
      ]<\/p>\n\n\n\n

      Thus, the foci<\/strong> are at:<\/p>\n\n\n\n

      [
      (h, k \\pm c) = \\left(-1, \\frac{3}{2} \\pm 3\\right)
      ]<\/p>\n\n\n\n

      So:<\/p>\n\n\n\n

        \n
      • Upper focus: ( \\left(-1, \\frac{9}{2}\\right) )<\/li>\n\n\n\n
      • Lower focus: ( \\left(-1, \\frac{-3}{2}\\right) )<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Step 2: Find endpoints of the latera recta:<\/h4>\n\n\n\n

        Since the major axis is vertical, the latus rectum at each focus is horizontal<\/strong>, passing through a focus.<\/p>\n\n\n\n

        Substitute ( y = \\frac{9}{2} ) into the ellipse equation to find corresponding ( x )-values:<\/p>\n\n\n\n

        Start with:<\/p>\n\n\n\n

        [
        \\frac{(x+1)^2}{4} + \\frac{(y-\\frac{3}{2})^2}{13} = 1
        ]<\/p>\n\n\n\n

        At ( y = \\frac{9}{2} ):<\/p>\n\n\n\n

        [
        \\frac{(x+1)^2}{4} + \\frac{\\left(\\frac{9}{2} – \\frac{3}{2}\\right)^2}{13} = 1
        ]
        Simplify:<\/p>\n\n\n\n

        [
        \\frac{(x+1)^2}{4} + \\frac{(3)^2}{13} = 1
        ]
        [
        \\frac{(x+1)^2}{4} + \\frac{9}{13} = 1
        ]
        [
        \\frac{(x+1)^2}{4} = \\frac{4}{13}
        ]
        [
        (x+1)^2 = \\frac{16}{13}
        ]
        [
        x+1 = \\pm \\frac{4}{\\sqrt{13}}
        ]
        Thus:<\/p>\n\n\n\n

        [
        x = -1 \\pm \\frac{4}{\\sqrt{13}}
        ]<\/p>\n\n\n\n

        Similarly, at ( y = -\\frac{3}{2} ) (lower focus):<\/p>\n\n\n\n

        Substitute ( y = -\\frac{3}{2} ) into the ellipse:<\/p>\n\n\n\n

        [
        \\frac{(x+1)^2}{4} + \\frac{\\left(-\\frac{3}{2} – \\frac{3}{2}\\right)^2}{13} = 1
        ]
        Simplify:<\/p>\n\n\n\n

        [
        \\frac{(x+1)^2}{4} + \\frac{(-3)^2}{13} = 1
        ]
        [
        \\frac{(x+1)^2}{4} + \\frac{9}{13} = 1
        ]
        [
        \\frac{(x+1)^2}{4} = \\frac{4}{13}
        ]
        Thus the same computation for ( x ):<\/p>\n\n\n\n

        [
        x = -1 \\pm \\frac{4}{\\sqrt{13}}
        ]<\/p>\n\n\n\n

        \n\n\n\n

        Final Answer:<\/h4>\n\n\n\n
          \n
        • Endpoints of upper latus rectum:
          [
          \\left( -1 + \\frac{4}{\\sqrt{13}}, \\frac{9}{2} \\right) \\quad \\text{and} \\quad \\left( -1 – \\frac{4}{\\sqrt{13}}, \\frac{9}{2} \\right)
          ]<\/li>\n\n\n\n
        • Endpoints of lower latus rectum:
          [
          \\left( -1 + \\frac{4}{\\sqrt{13}}, -\\frac{3}{2} \\right) \\quad \\text{and} \\quad \\left( -1 – \\frac{4}{\\sqrt{13}}, -\\frac{3}{2} \\right)
          ]<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Short Explanation (About 300 words)<\/strong><\/h3>\n\n\n\n

          The latus rectum<\/strong> of an ellipse is a chord that passes through a focus and is perpendicular to the major axis<\/strong>. In this problem, we were given a conic section in a general quadratic form. We first rewrote it<\/strong> into standard form<\/strong> by completing the square. After completing the square, we obtained the standard ellipse equation, recognizing that the major axis is vertical<\/strong> because ( b^2 > a^2 ).<\/p>\n\n\n\n

          From the standard form, we identified the center ( (-1, 3\/2) ) and semi-axes lengths ( a=2 ) and ( b=\\sqrt{13} ). To find the foci, we computed the focal distance ( c = 3 ), and so the foci are located vertically above and below the center.<\/p>\n\n\n\n

          To find the endpoints of the latus recta, we substituted the ( y )-coordinates of the foci into the ellipse equation and solved for corresponding ( x )-values. Since the latus rectum is horizontal<\/strong> in this case (vertical major axis), the endpoints are horizontally displaced from the focus. Specifically, at each focus, the two endpoints are ( \\frac{8}{\\sqrt{13}} ) units apart horizontally.<\/p>\n\n\n\n

          Thus, the endpoints were expressed exactly in terms of square roots: ( -1 \\pm \\frac{4}{\\sqrt{13}} ) for ( x ), with corresponding ( y )-values ( \\frac{9}{2} ) and ( -\\frac{3}{2} ). This detailed and exact method ensures full understanding of the geometric properties of ellipses.<\/p>\n\n\n\n

          \n","protected":false},"excerpt":{"rendered":"

          A latus rectum of an ellipse is a chord passing through the foci and perpendicular to the major axis. Suppose that an ellipse has major and minor axes of lengths 2a and 2b, where a>b. a) (3 marks) Show that the length of a latus rectum of an ellipse is 2b2\/a. Hint: The foci of […]<\/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-208442","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208442","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=208442"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208442\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208442"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208442"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}