{"id":3260,"date":"2021-07-21T17:01:58","date_gmt":"2021-07-21T17:01:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3260"},"modified":"2021-11-23T21:40:24","modified_gmt":"2021-11-23T16:10:24","slug":"equation-of-tangent-to-ellipse-in-all-forms","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-tangent-to-ellipse-in-all-forms\/","title":{"rendered":"Equation of Tangent to Ellipse in all Forms"},"content":{"rendered":"
The equation of tangent to the given ellipse at its point (\\(x_1, y_1\\)) is\u00a0<\/p>\n
\n\\(x{x_1}\\over a^2\\) + \\(y{y_1}\\over b^2\\) = 1.<\/p>\n<\/blockquote>\n
Note – <\/strong>For general ellipse replace \\(x^2\\) by \\(xx_1\\), \\(y^2\\) by \\(yy_1\\), 2x by \\(x + x_1\\), 2y by \\(y + y_1\\), 2xy by \\(xy_1 + yx_1\\) and c by c.<\/p>\n
(b) Slope form :<\/h2>\n
The Equation of tangent to the given ellipse whose slope is ‘m<\/strong>‘, is<\/p>\n
\ny = mx \\(\\pm\\) \\(\\sqrt{a^2m^2 + b^2}\\),<\/p>\n
Point of contact are (\\({\\pm} a^2m\\over \\sqrt{a^2m^2 + b^2}\\), \\({\\pm} b^2\\over \\sqrt{a^2m^2 + b^2}\\)).<\/p>\n<\/blockquote>\n
Note that there are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction.<\/p>\n
(c) Parametric form :<\/h2>\n
The equation of tangent to the given ellipse at its point (acos\\(\\theta\\), bsin\\(\\theta\\))<\/strong>, is<\/p>\n
\n\\(xcos\\theta\\over a\\) + \\(ysin\\theta\\over b\\) = 1<\/p>\n<\/blockquote>\n
Note<\/strong> – The point of the intersection of the tangents at the point \\(\\alpha\\) & \\(\\beta\\) is (a\\(cos{\\alpha+\\beta\\over 2}\\over cos{\\alpha -\\beta\\over 2}\\), b\\(sin{\\alpha+\\beta\\over 2}\\over cos{\\alpha-\\beta\\over 2}\\))<\/p>\n\n\n
Example : <\/span> Find the equation of the tangents to the ellipse \\(3x^2+4y^2\\) = 12 which are perpendicular to the line y + 2x = 4<\/p>\n
Solution : <\/span>Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4
\\(\\therefore\\) mx – 2 = -1 \\(\\implies\\) m = \\(1\\over 2\\)
Since \\(3x^2+4y^2\\) = 12 or \\(x^2\\over 4\\) + \\(y^2\\over 3\\) = 1
Comparing this with \\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1
\\(\\therefore\\) \\(a^2\\) = 4 and \\(b^2\\) = 3
So the equation of the tangent are y = \\(1\\over 2\\)x \\(\\pm\\) \\(\\sqrt{4\\times {1\\over 4} + 3}\\)
\\(\\implies\\) y = \\(1\\over 2\\)x \\(\\pm\\) 2 or x – 2y \\(\\pm\\) 4 = 0
<\/p>\n\n\nHope you learnt equation of tangent 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