{"id":3269,"date":"2021-07-21T17:01:53","date_gmt":"2021-07-21T17:01:53","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3269"},"modified":"2021-11-23T21:41:55","modified_gmt":"2021-11-23T16:11:55","slug":"equation-of-normal-to-ellipse-in-all-forms","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-normal-to-ellipse-in-all-forms\/","title":{"rendered":"Equation of Normal to Ellipse in all Forms"},"content":{"rendered":"
The Equation of normal to the given ellipse at (\\(x_1, y_1\\)) is<\/p>\n
\n\\(a^2x\\over x_1\\) + \\(b^2y\\over y_1\\) = \\(a^2-b^2\\) = \\(a^2e^2\\)<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Find the normal to the ellipse \\(9x^2+16y^2\\) = 288 at the point (4,3).<\/p>\n
Solution : <\/span>We have, \\(9x^2+16y^2\\) = 288
\nComparing with general equation of ellipse,
\n\\(a^2\\) = 32 and \\(b^2\\) = 18
\nThe normal to given ellipse in point form is \\(a^2x\\over x_1\\) + \\(b^2y\\over y_1\\) = \\(a^2-b^2\\)
\nSo, \\(32x\\over 4\\) + \\(18y\\over 3\\) = 14
\n 8x + 6y = 14
\nHence, normal to given ellipse is 8x + 6y = 14.<\/p>\n\n\n(b) Slope form :<\/h3>\n
The equation of normal to the given ellipse whose slope is ‘m’, is<\/p>\n
\ny = mx \\(\\mp\\) \\({(a^2-b^2)m}\\over \\sqrt{a^2 + b^2m^2}\\)<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Find the normal to the ellipse \\(x^2 + 2y^2\\) = 6 whose slope is 2.<\/p>\n
Solution : <\/span>We have, \\(x^2 + 2y^2\\) = 6
\nComparing with \\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1
\n\\(a^2\\) = 6 and \\(b^2\\) = 3
\nThe normal to given ellipse in slope form is y = mx \\(\\mp\\) \\({(a^2-b^2)m}\\over \\sqrt{a^2 + b^2m^2}\\)
\nSo, y = 2x \\(\\mp\\) \\({3m}\\over \\sqrt{18}\\)
\nHence, normal to given ellipse isy = 2x \\(\\mp\\) \\({3m}\\over \\sqrt{18}\\).<\/p>\n\n\n(c) Parametric form :<\/h3>\n
The equation of normal to the given ellipse at its point (acos\\(\\theta\\), bsin\\(\\theta\\)), is<\/p>\n
\n\\(axsec\\theta \\) – \\(bycosec\\theta \\) = \\(a^2-b^2\\)<\/p>\n
or \\(ax\\over cos\\theta\\) – \\(by\\over sin\\theta\\) = \\(a^2-b^2\\)<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Find the condition that the line lx + my = n may be a normal to the ellipse \\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1<\/p>\n
Solution : <\/span>Equation of normal to the given ellipse at its point (acos\\(\\theta\\), bsin\\(\\theta\\)), is
\n\\(ax\\over cos\\theta\\) – \\(by\\over sin\\theta\\) = \\(a^2-b^2\\) ….(i)
\nIf line lx + my = n is also normal to the ellipse then there must be a value of \\(\\theta\\) for which line (i) and line lx + my = n are identical. for that value of \\(\\theta\\) we have
\n\\(l\\over ({a\\over cos\\theta})\\) = \\(m\\over -({b\\over sin\\theta})\\) = \\(n\\over {l(a^2 – b^2)}\\) or
\n\\(\\cos\\theta\\) = \\(an\\over {l(a^2 – b^2)}\\) ……(iii)
\n\\(\\sin\\theta\\) = \\(-bn\\over {m(a^2 – b^2)}\\) ……(iv)
\nSquaring and adding (iii) and (iv), we get\n\\(n^2\\over {(a^2 – b^2)^2}\\) (\\({a^2\\over l^2} + {b^2\\over m^2}\\)) which is the required condition.<\/p>\n\n\nHope you learnt equation of normal to ellipse in all forms, learn more concepts of ellipse and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\n