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
8.1 Point form
7.2 Parametric form
7.3 Equation of tangent in terms of slope/Slope form
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 $$\frac{{{a^2}x}}{{{x_1}}} - \frac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}$$
Proof: Since the equation of tangent to ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at point $({x_1},{y_1})$ is $$\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1$$
The slope of tangent at $({x_1},{y_1})$ $ = - \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}$.
Therefore, the slope of normal at $({x_1},{y_1})= - \frac{1}{{slope\ of\operatorname{tangent} }} = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}$.
Hence the equation of normal at $({x_1},{y_1})$ is $$y - {y_1} = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}(x - {x_1})$$
or, $$\frac{{{a^2}x}}{{{x_1}}} - \frac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}$$
Question 10. If the normal at the end of a latus rectum of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ passes through the extremity of a minor axis, prove that ${e^4} + {e^2} - 1 = 0$.
Solution: As we know that the end of latus rectum is $P(ae,\frac{{{b^2}}}{a})$, therefore the equation of normal at a point $P$ using point form is $$y - {y_1} = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}(x - {x_1})$$ or, $$\frac{{{a^2}(x - ae)}}{{ae}} = \frac{{a{b^2}(y - \frac{{{b^2}}}{a})}}{{{b^2}}}...(1)$$
Since it passes through extremity of a minor axis i.e., $(0,-b)$ which satisfies the equation $(1)$,
$$ - {a^2} = - ab - {b^2}$$ $${a^2} = ab + {a^2}(1 - {e^2}){\text{ [}}\because {b^2} = {a^2}(1 - {e^2})]$$ or, $$b = a{e^2}$$
Squaring both sides we get, $${b^2} = {a^2}{e^4}$$
Again put ${b^2} = {a^2}(1 - {e^2})$, we get $${e^4} + {e^2} - 1 = 0$$
