QUESTION

# If the normal at the end of the latus rectum of ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ passes through$(0, - b)$, then find the value of ${e^4} + {e^2}$(where $e$ is eccentricity).

Hint: We know that the coordinates of the endpoint of latus rectum of ellipse lying in the first quadrant is $\left( {ae,\dfrac{{{b^2}}}{a}} \right)$. Now, according to the information given in the question, a normal is drawn at this point of ellipse.

The general equation of normal to ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ at point $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{{a^2}x}}{{{x_1}}} - \dfrac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}$. Here, the point at which normal is drawn is $\left( {ae,\dfrac{{{b^2}}}{a}} \right)$. So, equation of normal is:

$\Rightarrow \dfrac{{{a^2}x}}{{ae}} - \dfrac{{{b^2}y}}{{\dfrac{{{b^2}}}{a}}} = {a^2} - {b^2}, \\ \Rightarrow \dfrac{{ax}}{e} - ay = {a^2} - {b^2} \\$
Now, again from the question, the normal is passing through $(0, - b)$ therefore, satisfying this point in the above equation, we’ll get:
$\Rightarrow \dfrac{{a \times 0}}{e} - a( - b) = {a^2} - {b^2}, \\ \Rightarrow ab = {a^2} - {b^2} .....(i) \\$
For ellipse, we know that:
$\Rightarrow {b^2} = {a^2}(1 - {e^2}) .....(ii)$
$\Rightarrow {a^2} - {b^2} = {a^2}{e^2}$
Putting this in equation$(i)$, we’ll get:
$\Rightarrow {a^2}{e^2} = ab, \\ \Rightarrow {e^2} = \dfrac{b}{a} \\$
From equation$(ii)$, we have:
$\Rightarrow {b^2} = {a^2}(1 - {e^2}) \\ \Rightarrow {\left( {\dfrac{b}{a}} \right)^2} = (1 - {e^2}) \\$
Putting the value of $\dfrac{b}{a}$ from above:
$\Rightarrow {({e^2})^2} = 1 - {e^2}, \\ \Rightarrow {e^4} = 1 - {e^2}, \\ \Rightarrow {e^4} + {e^2} = 1 \\$
Thus the value of ${e^4} + {e^2}$ is $1$.

Note: The equation of normal can also be determined using point-slope method. The point of normality is the end point of the latus rectum and the slope of normal is the value of $- \dfrac{{dx}}{{dy}}$ at that point. So, for this method we have to calculate the value of $- \dfrac{{dx}}{{dy}}$ separately.