If the normal at an end of latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $$e$$ of the ellipse satisfies:

Let us take the ellipse in its standard (centre-origin, major axis on the X-axis) form

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b>0.$$

For this ellipse we recall the following facts:

• The distance of each focus from the centre is $$c,$$ where $$c^{2}=a^{2}-b^{2}.$$

• The eccentricity is defined by $$e=\dfrac{c}{a},$$ so that $$e^{2}=\dfrac{c^{2}}{a^{2}}=\dfrac{a^{2}-b^{2}}{a^{2}}=1-\dfrac{b^{2}}{a^{2}}.$$

• The latus rectum corresponding to the focus $$(c,0)$$ is the line $$x=c,$$ and its two end-points on the ellipse are

$$\left(c,\;\;\frac{b^{2}}{a}\right)\quad\text{and}\quad\left(c,\;-\frac{b^{2}}{a}\right).$$

We choose the upper end of the latus rectum, namely

$$P\bigl(c,\;y_{1}\bigr)=\left(c,\;\frac{b^{2}}{a}\right).$$

The slope of the tangent at any point $$(x,y)$$ of the ellipse is obtained by implicit differentiation of the defining equation:

$$\frac{d}{dx}\!\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=\frac{d}{dx}(1) \;\Longrightarrow\;\frac{2x}{a^{2}}+\frac{2y}{b^{2}}\frac{dy}{dx}=0,$$

hence

$$\frac{dy}{dx}=m_{t}=-\frac{b^{2}x}{a^{2}y}.$$

Therefore the slope of the normal is the negative reciprocal:

$$m_{n}=+\frac{a^{2}y}{b^{2}x}.$$

Evaluating this at the point $$P(c,\;b^{2}/a)$$ we get

$$m_{n}=\frac{a^{2}\,(b^{2}/a)}{b^{2}\,c}=\frac{a}{c}.$$

Thus the equation of the normal at $$P$$ is

$$y-\frac{b^{2}}{a}=\frac{a}{c}\Bigl(x-c\Bigr).$$

According to the statement of the problem, this normal line passes through an extremity of the minor axis. The extremities of the minor axis are $$(0,b)$$ and $$(0,-b).$$ Substituting $$(0,-b)$$ (we shall see that this choice gives a consistent, positive value of $$b$$) in the equation of the normal, we have

$$-b-\frac{b^{2}}{a}=\frac{a}{c}\bigl(0-c\bigr)=\frac{a}{c}(-c)=-a.$$

Collecting like terms:

$$-b-\frac{b^{2}}{a}+a=0.$$

Multiplying every term by $$a$$ to clear the denominator,

$$-ab-b^{2}+a^{2}=0,$$

or equivalently

$$a^{2}-ab-b^{2}=0.$$

Now we wish to express everything in terms of the single variable $$e.$$ For this purpose set

$$k=\frac{b}{a}\quad\bigl(\;0<k<1\;\bigr).$$

Dividing the last obtained relation by $$a^{2}$$ gives

$$1-k-k^{2}=0,$$

which can be written as

$$k^{2}+k-1=0.$$

Next we relate $$k$$ to $$e.$$ From the basic identity quoted earlier,

$$e^{2}=1-\frac{b^{2}}{a^{2}}=1-k^{2}.$$

Because $$k^{2}=1-k$$ from the quadratic immediately above, we substitute:

$$e^{2}=1-(1-k)=k.$$

Thus we have obtained

$$k=e^{2}.$$

Returning to the quadratic satisfied by $$k,$$ namely $$k^{2}+k-1=0,$$ we now replace $$k$$ by $$e^{2}:$$

$$\bigl(e^{2}\bigr)^{2}+e^{2}-1=0.$$

Simplifying,

$$e^{4}+e^{2}-1=0.$$

Hence, the correct answer is Option C.