{"id":8107,"date":"2021-11-13T23:11:32","date_gmt":"2021-11-13T17:41:32","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8107"},"modified":"2021-11-14T00:34:38","modified_gmt":"2021-11-13T19:04:38","slug":"find-the-slope-of-normal-to-the-curve-x-1-a-sintheta-y-b-cos2theta-at-theta-piover-2","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-slope-of-normal-to-the-curve-x-1-a-sintheta-y-b-cos2theta-at-theta-piover-2\/","title":{"rendered":"Find the slope of normal to the curve x = 1 – \\(a sin\\theta\\), y = \\(b cos^2\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 2\\)."},"content":{"rendered":"
We have, x = 1 – \\(a sin\\theta\\), y = \\(b cos^2\\theta\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(dx\\over d\\theta\\) = \\(-a cos\\theta\\)\u00a0 and \\(dy\\over d\\theta\\) = \\(-2b cos\\theta sin\\theta\\)<\/p>\n
\\(\\therefore\\)\u00a0 \\(dy\\over dx\\) = \\(dy\/d\\theta\\over dx\/d\\theta\\) = \\(2b\\over a\\) \\(sin\\theta\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(dy\\over dx\\)\u00a0 at \\(\\pi\\over 2\\) = \\(2b\\over a\\)<\/p>\n
Hence, Slope of normal<\/a> at \\(\\theta\\) = \\(\\pi\\over 2\\) = \\(-a\\over 2b\\)<\/p>\n Find the slope of the normal to the curve x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 4\\).<\/a><\/p>\n Show that the tangents to the curve y = \\(2x^3 \u2013 3\\) at the points where x =2 and x = -2 are parallel.<\/a><\/p>\n Find the equations of the tangent and the normal at the point \u2018t\u2019 on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).<\/a><\/p>\n Find the equation of the normal to the curve y = \\(2x^2 + 3 sin x\\) at x = 0.<\/a><\/p>\n
\nSimilar Questions<\/h3>\n