{"id":174741,"date":"2018-02-20T23:50:30","date_gmt":"2018-02-20T23:50:30","guid":{"rendered":"https:\/\/writemyessayfree.com\/how-do-you-find-the-point-in-the-first-quadrant-on-the-lemniscate-2x2y2225x2%e2%88%92y2-where-the-tangent-is-horizontal"},"modified":"2018-10-24T09:24:47","modified_gmt":"2018-10-24T09:24:47","slug":"how-do-you-find-the-point-in-the-first-quadrant-on-the-lemniscate-2x2y2225x2%e2%88%92y2-where-the-tangent-is-horizontal","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/how-do-you-find-the-point-in-the-first-quadrant-on-the-lemniscate-2x2y2225x2%e2%88%92y2-where-the-tangent-is-horizontal\/","title":{"rendered":"How do you find the point (in the first quadrant) on the lemniscate ##2(x^2+y^2)2=25(x^2\u2212y^2)## where the tangent is horizontal?"},"content":{"rendered":"<p>The point on the lemniscate that has a horizontal tangent, in the first quadrant, is ##((5sqrt(3))\/4, 5\/4)##. Refer to the explanation<\/p>\n<p>The lemniscate equation is ##2(x^2 + y^2)^2 = 25(x^2 &#8211; y^2)## Here are a few things you can extract from the question. I&#8217;ll refer to this information when necessary (you&#8217;ll know when):<\/p>\n<ul>\n<li>##color(orange)[&#8220;a horizontal tangent line means that the slope is zero&#8221;]##. Mathematically, ##y&#8217;= (dy)\/dx = 0##.<\/li>\n<li>##color(red)[&#8220;first quadrant answers mean that all x and y values are positive&#8221;]##<\/li>\n<li>##color(blue)[&#8220;you&#8217;ll need to implicitly differentiate&#8221;]##. Let ##y&#8217; = (dy)\/dx##.<\/li>\n<\/ul>\n<p>Step 1) The requires the and the . ##2(x^2 + y^2)^2 = 25(x^2 &#8211; y^2)## ##2(2)(x^2 +y^2)^1(2x + color(blue)(2yy&#8217;)) = 25(2x-color(blue)(2yy&#8217;))## ##4(x^2 +y^2)(2x + 2yy&#8217;) = 25(2x-2yy&#8217;)## ##4(x^2+y^2)(2x + color(orange)0) = 25(2x &#8211; color(orange)0)## ##4(x^2+y^2)(2x) = 25(2x)## ##4(x^2 + y^2) = 25## ##x^2 + y^2 = 25\/4##<\/p>\n<p>Step 2) You can solve for either ##x## or ##y## to find the values of the coordinate. I&#8217;ll solve for ##x##. ##x^2 + y^2 = 25\/4## ##x^2 = 25\/4 &#8211; y^2## ##x = +-sqrt(25\/4 &#8211; y^2)##. Since the point lies in the first quadrant, ##color(red)[x = sqrt(25\/4-y^2)]##<\/p>\n<p>Step 3) Back to the original equation to solve for y: ##2(x^2 + y^2)^2 = 25(x^2 &#8211; y^2)## ##2([sqrt(25\/4-y^2)]^2 + y^2)^2 = 25([sqrt(25\/4-y^2)]^2 &#8211; y^2)## ##2(25\/4-y^2 + y^2)^2 = 25(25\/4-y^2 &#8211; y^2)## ##2(25\/4)^2 = 25(25\/4 &#8211; 2y^2)## ##(2*25*25)\/(4*4*25) = 25\/4 &#8211; 2y^2## ##25\/8 = 25\/4 &#8211; 2y^2## ##2y^2 = 25\/4 &#8211; 25\/8## ##16y^2 = 50 &#8211; 25## ##16y^2 = 25## ##y^2 = 25\/16## ##y = +- sqrt(25\/16)## ##color(red)[y = sqrt(25\/16)]## ##y = 5\/4##<\/p>\n<p>Step 4) Back to the statement ##x = sqrt(25\/4-y^2)##, ##x = sqrt(25\/4-(5\/4)^2)## ##x = sqrt(25\/4-25\/16)## ##x = sqrt((100-75)\/16)## ##x = +-sqrt(75\/16)## ##color(red)[x = (5sqrt(3))\/4]##<\/p>\n<p>The point on the lemniscate that has a horizontal tangent, in the first quadrant, is ##((5sqrt(3))\/4, 5\/4)##.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The point on the lemniscate that has a horizontal tangent, in the first quadrant, is ##((5sqrt(3))\/4, 5\/4)##. Refer to the explanation The lemniscate equation is ##2(x^2 + y^2)^2 = 25(x^2 &#8211; y^2)## Here are a few things you can extract <a href=\"https:\/\/www.benedictsol.com\/blogs\/how-do-you-find-the-point-in-the-first-quadrant-on-the-lemniscate-2x2y2225x2%e2%88%92y2-where-the-tangent-is-horizontal\/\" class=\"read-more\">Read More &#8230;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-174741","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/174741","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=174741"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/174741\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=174741"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=174741"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=174741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}