{"id":3322,"date":"2021-07-22T11:33:46","date_gmt":"2021-07-22T11:33:46","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3322"},"modified":"2021-11-27T18:14:07","modified_gmt":"2021-11-27T12:44:07","slug":"equation-of-tangent-to-parabola","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-tangent-to-parabola\/","title":{"rendered":"Equation of Tangent to Parabola in all Forms"},"content":{"rendered":"
The equation of tangent to parabola in point form, slope form and parametric form are given below with examples.<\/p>\n
(a)\u00a0 The line y = mx + c meets the parabola \\(y^2\\) = 4ax in two points real, coincident or imaginary according as a >=< cm \\(\\implies\\) condition of tangency<\/strong> is,<\/p>\n \u00a0Condition of tangency is \u00a0c = \\(a\\over m\\)<\/p>\n<\/blockquote>\n Note :<\/strong> Line y = mx + c will be tangent to parabola \\(x^2\\) = 4ay if c = -a\\(m^2\\)<\/p>\n (b) Length of the chord intercepted by the parabola \\(y^2\\) = 4ax on the line y = mx + c is \\(4\\over m^2\\)\\(\\sqrt{a(1+m^2)(a-mc)}\\)<\/p>\n Note :<\/strong> Length of the focal chord making an angle \\(\\alpha\\) with the x-axis is 4a \\({cosec}^2\\alpha\\).<\/p>\n The equation of tangent to the given parabola at its point (\\(x_1, y_1\\))<\/strong> is<\/p>\n y\\(y_1\\) = 2a(x + \\(x_1\\))<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the tangent to the parabola \\(y^2 = 16x\\) at (5, 2).<\/p>\n Solution : <\/span>We have, \\(y^2 = 16x\\) The equation of tangent to the given parabola whose slope is ‘m’, is<\/p>\n y = mx + \\(a\\over m\\), (m \\(\\ne\\) 0)<\/p>\n The point of contact is (\\(a\\over m^2\\), \\(2a\\over m\\))<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the tangent to the parabola \\(y^2 = 8x\\) whose slope is 3.<\/p>\n Solution : <\/span>We have, \\(y^2 = 8x\\) The tangent to the given parabola at its point P(t), is<\/p>\n ty = x + a\\(t^2\\)<\/p>\n Note –<\/strong> Point of intersection of the tangents at the points \\(t_1\\) & \\(t_2\\) is [a\\(t_1t_2\\), a(\\(t_1 + t_2\\))].<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the equation of the tangents to the parabola \\(y^2\\) = 9x which go through the point (4,10).<\/p>\n Solution : <\/span>tangent to the parabola \\(y^2\\) = 9x is Hope you learnt equation of tangent to parabola in point form, slope form and parametric form, learn more concepts of parabola <\/a><\/span>and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\n\n
Equation of Tangent to Parabola \\(y^2 = 4ax\\)<\/h2>\n
(a) Point form :<\/h3>\n
\n
\nCompare given equation with \\(y^2 = 4ax\\)
\n\\(\\implies\\) a = 4
\nHence, required equation of tangent is 2y = 8(x + 5)
\n= 2y = 8x + 40
<\/p>\n\n\n(b) Slope form :<\/h3>\n
\n
\nCompare given equation with \\(y^2 = 4ax\\)
\na = 2
\nHence, required equation of tangent is y = 3x + \\(2\\over 3\\)
<\/p>\n\n\n(c) Parametric form :<\/h3>\n
\n
\ny = mx + \\(9\\over 4m\\)
\nSince it passes through (4,10)
\n\\(\\therefore\\) 10 = 4m + \\(9\\over 4m\\) \\(\\implies\\) 16\\(m^2\\) – 40m + 9 = 0
\nm = \\(1\\over 4\\), \\(9\\over 4\\)
\n\\(\\therefore\\) Equation of tangent’s are y = \\(x\\over 4\\) + 9 & y = \\(9x\\over 4\\) + 1
<\/p>\n\n\n