Ellipse
1.0 Definition
2.0 Standard equation of Ellipse
3.0 Important terms
4.0 Difference between two forms of Ellipse
5.0 Focal Distance of a point
6.0 Parametric Co-ordinates
7.0 Equation of Tangent to Ellipse
7.1 Equation of tangent to a point/Point form
7.2 Parametric form
7.3 Equation of tangent in terms of slope/Slope form
8.0 Equation of Normal to Ellipse
9.0 Pair of tangents
10.0 Chord of contact
11.0 Chord bisected at a given point
12.0 Director circle
8.2 Parametric form
7.2 Parametric form
7.3 Equation of tangent in terms of slope/Slope form
If the parametric coordinates of a point at which normal is drawn is given, then the equation of normal at a point $(a\cos \theta ,b\sin \theta )$ to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is $$ax\sec \theta - by\operatorname{cosec} \theta = {a^2} - {b^2}$$
Proof: Since the equation of normal at a point $({x_1},{y_1})$ to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is $$\frac{{{a^2}x}}{{{x_1}}} - \frac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}...(1)$$
Replacing ${x_1}$ by $a\cos \theta $ and ${y_1}$ by $b\sin \theta $ in equation $(1)$, we get $$\frac{{{a^2}x}}{{a\cos \theta }} - \frac{{{b^2}y}}{{b\sin \theta }} = {a^2} - {b^2}$$
or, $$ax\sec \theta - by\operatorname{cosec} \theta = {a^2} - {b^2}$$
Question 11. Prove that the straight line $lx + my + n = 0$ is a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ if $$\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$$
Solution: The equation of normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ in parametric form is $$ax\sec \theta - by\operatorname{cosec} \theta = {a^2} - {b^2}...(1)$$
As it is given in the question that the staright line $lx + my + n = 0...(2)$ is a normal to the ellipse, therefore, the equation $(1)$ and $(2)$ are same equations. On comparing both the equations we get, $$\frac{{a\sec \theta }}{l} = - \frac{{b\operatorname{cosec} \theta }}{m} = \frac{{{a^2} - {b^2}}}{{ - n}}$$
Therefore, $\cos \theta = - \frac{{na}}{{l({a^2} - {b^2})}}...(3)$ and $\sin \theta = \frac{{nb}}{{m({a^2} - {b^2})}}...(4)$
Squaring and adding equations $(3)$ and $(4)$, we get
$$\begin{equation} \begin{aligned} \frac{{{n^2}{b^2}}}{{{m^2}{{({a^2} - {b^2})}^2}}} + \frac{{{n^2}{a^2}}}{{{l^2}{{({a^2} - {b^2})}^2}}} = 1 \\ \frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}} \\\end{aligned} \end{equation} $$
