Equation of Tangent to Ellipse in all Forms

(a) Point form :

The equation of tangent to the given ellipse at its point ($$x_1, y_1$$) is

$$x{x_1}\over a^2$$ + $$y{y_1}\over b^2$$ = 1.

Note – 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 ‘m‘, 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 – 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

$$\therefore$$   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

Hope 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!