{"id":5593,"date":"2021-09-22T16:09:05","date_gmt":"2021-09-22T10:39:05","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5593"},"modified":"2021-11-13T21:35:53","modified_gmt":"2021-11-13T16:05:53","slug":"equation-of-tangent-and-normal-to-the-curve","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-tangent-and-normal-to-the-curve\/","title":{"rendered":"Equation of Tangent and Normal to the Curve"},"content":{"rendered":"
Here you will learn equation of tangent and normal to the curve with examples.<\/p>\n
Let’s begin –<\/p>\n
We know that the equation of line passing through a point \\((x_1, y_1)\\)<\/strong> and having slope m is \\(y – y_1\\) = m\\((x – x_1)\\).<\/strong><\/p>\n and we know that the slopes of the tangent and the normal to the curve y = f(x) at a point P\\((x_1, y_1)\\) are \\(({dy\\over dx})_P\\)<\/strong> and -\\(1\\over ({dy\\over dx})_P\\)<\/strong> respectively. <\/p>\n Therefore the equation of the tangent <\/strong>at P\\((x_1, y_1)\\) to the curve y = f(x) is<\/p>\n \\(y – y_1\\) = \\(({dy\\over dx})_P\\) (\\(x – x_1\\))<\/p>\n<\/blockquote>\n Since the normal at P\\((x_1, y_1)\\) passes through P and has slope -\\(1\\over ({dy\\over dx})_P\\).<\/p>\n Therefore, the equation of the normal<\/strong> at P\\((x_1, y_1)\\) to the curve y = f(x) is<\/p>\n \\(y – y_1\\) = \\(-1\\over ({dy\\over dx})_P\\) (\\(x – x_1\\))<\/p>\n<\/blockquote>\n Note :<\/strong><\/p>\n 1). If \\(({dy\\over dx})_P\\) = \\(\\pm \\infty\\), then the tangent at \\((x_1, y_1)\\) is parallel to y-axis and its equation is x = \\(x_1\\).<\/p>\n 2). If \\(({dy\\over dx})_P\\) = 0, then the normal at \\((x_1, y_1)\\) is parallel to y-axis and its equation is x = \\(x_1\\).<\/p>\n 3). If \\(({dy\\over dx})_P\\) = \\(\\pm \\infty\\), then the normal at \\((x_1, y_1)\\) is parallel to x-axis and its equation is y = \\(y_1\\).<\/p>\n 4). If \\(({dy\\over dx})_P\\) = 0, then the tangent at \\((x_1, y_1)\\) is parallel to x-axis and its equation is y = \\(y_1\\).<\/p>\n Example<\/span><\/strong> : find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\) at the point (1\/2, 35\/4).<\/p>\n Solution<\/span><\/strong> : The equation of the given curve is <\/p>\n y = \\(-5x^2 + 6x + 7\\) <\/p>\n \\(\\implies\\) \\(dy\\over dx\\) = -10x + 6<\/p>\n \\(\\implies\\) \\(({dy\\over dx})_{(1\/2, 35\/4)}\\) = \\(-10\\over 4\\) + 6 = 1<\/p>\n The required equation at (1\/2, 35\/4) is<\/p>\n y – \\(35\\over 4\\) = \\(({dy\\over dx})_{(1\/2, 35\/4)}\\) \\((x – {1\\over 2})\\)<\/p>\n \\(\\implies\\) y – 35\/4 = 1(x – 1\/2)<\/p>\n \\(\\implies\\) y = x + 33\/4<\/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\n\n\n
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\nRelated Questions<\/h3>\n