{"id":3332,"date":"2021-07-22T11:34:12","date_gmt":"2021-07-22T11:34:12","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3332"},"modified":"2021-11-27T18:15:32","modified_gmt":"2021-11-27T12:45:32","slug":"what-is-the-equation-of-normal-to-parabola","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-equation-of-normal-to-parabola\/","title":{"rendered":"Equation of Normal to Parabola in all Forms"},"content":{"rendered":"
The equation of normal to parabola in point form, slope form and parametric form are given below with examples.<\/p>\n
The equation of normal to the given parabola at its point (\\(x_1, y_1\\))<\/strong> is<\/p>\n y – \\(y_1\\) = \\(-y_1\\over 2a\\)(x – \\(x_1\\))<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the normal to the parabola \\(y^2 = 32x\\) at (3, 1).<\/p>\n Solution : <\/span>We have, \\(y^2 = 32x\\) The equation of normal to the given parabola whose slope is ‘m’, is <\/p>\n y = mx – 2am – a\\(m^3\\)<\/p>\n The foot of the normal is (\\(am^2\\), -2am)<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the normal to the parabola \\(y^2 = 4x\\) whose slope is 3.<\/p>\n Solution : <\/span>We have, \\(y^2 = 4x\\) The normal to the given parabola at its point P(t), is<\/p>\n y + tx = 2at + a\\(t^3\\)<\/p>\n<\/blockquote>\n Note –<\/strong> Point of intersection of the normals at the points \\(t_1\\) & \\(t_2\\) is<\/p>\n [(a\\({t_1}^2 + {t_2}^2 + t_1t_2 + 2\\)), -a\\(t_1t_2(t_1 + t_2)\\)].<\/p>\n<\/blockquote>\n\n\n Example : <\/span> If two normals drawn from any point to the parabola \\(y^2\\) = 4ax make angle \\(\\alpha\\) and \\(\\beta\\) with the axis such that \\(tan\\alpha\\).\\(tan\\beta\\) = 2, then find the locus of this point.<\/p>\n Solution : <\/span>Let the point is (h, k). The equation of any normal to the parabola \\(y^2\\) = 4ax is Hope you learnt equation of normal 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
\nCompare given equation with \\(y^2 = 4ax\\)
\n\\(\\implies\\) a = 8
\nHence, required equation of normal is y – 1 = \\(-1\\over 16\\)(x – 3)
\n=> 16y + x = 19
<\/p>\n\n\n(b) Slope form<\/strong> :<\/h3>\n
\n
\nCompare given equation with \\(y^2 = 4ax\\)
\na = 1
\nHence, required equation of normal is y = 3x – 33
<\/p>\n\n\n(c) Parametric form<\/strong> :<\/h3>\n
\n
\n
\ny = mx – 2am – a\\(m^3\\)
\nSince it passes through (h,k)
\nk = mh – 2am – a\\(m^3\\)
\na\\(m^3\\) + m(2a – h) + k = 0 ……(i)
\n\\(m_1\\), \\(m_2\\) and \\(m_3\\) are roots of equation, then
\n\\(m_1\\).\\(m_2\\).\\(m_3\\) = \\(-k\\over a\\)
\nbut \\(m_1\\).\\(m_2\\) = 2, \\(m_3\\) = \\(-k\\over 2a\\)
\n\\(m_3\\) is the root of (i)
\n\\(\\therefore\\) \\(k^2\\) = 4ah
\nThus locus is \\(y^2\\) = 4ax.<\/p>\n\n\n