Equation of Tangent to Ellipse \(x^2\over a^2\) + \(y^2\over b^2\) = 1 :
(a) Point form :
The equation of tangent to the given ellipse at its point (\(x_1, y_1\)) isย
\(x{x_1}\over a^2\) + \(y{y_1}\over b^2\) = 1.
Note โ 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.
(b) Slope form :
The Equation of tangent to the given ellipse whose slope is โmโ, is
y = mx \(\pm\) \(\sqrt{a^2m^2 + b^2}\),
Point of contact are (\({\pm} a^2m\over \sqrt{a^2m^2 + b^2}\), \({\pm} b^2\over \sqrt{a^2m^2 + b^2}\)).
Note that there are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction.
(c) Parametric form :
The equation of tangent to the given ellipse at its point (acos\(\theta\), bsin\(\theta\)), is
\(xcos\theta\over a\) + \(ysin\theta\over b\) = 1
Note โ 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\(sin{\alpha+\beta\over 2}\over cos{\alpha-\beta\over 2}\))
Example : Find the equation of the tangents to the ellipse \(3x^2+4y^2\) = 12 which are perpendicular to the line y + 2x = 4
Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4
\(\therefore\) ย mx โ 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
\(\therefore\) ย \(a^2\) = 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 โ 2y \(\pm\) 4 = 0
Hope you learnt equation of tangent to ellipse in all forms, learn more concepts of ellipseย and practice more questions to get ahead in the competition. Good luck!
