Question:
<p>The normal at one of the ends of a latus rectum of the ellipse b<sup>2</sup>x<sup>2</sup> + a<sup>2</sup>y<sup>2</sup> = a<sup>2</sup>b<sup>2</sup> passes through the end of the minor axis. Then e satisfies</p>
<p style="display:inline">e<sup>2</sup> - e - 1 = 0</p>
<p style="display:inline">2e<sup>4</sup> + e<sup>2</sup> - 1 = 0</p>
<p style="display:inline">e<sup>4</sup> - e<sup>2</sup> - 1 = 0</p>
<p style="display:inline">e<sup>4</sup> + e<sup>2</sup> - 1 = 0</p>
Step-by-Step Solution
Key Concept: Establish the equation of the normal at the latus rectum extremity (ae, b²/a) and substitute the coordinates of the minor axis endpoint (0, -b) to relate eccentricity with the ellipse's axes.
<p><img alt="" height="104" src="https://i.imgur.com/eCSJm9V.png" width="112" /><br />
Let P be the end of a latus rectum.<br />
Equation of the tangent at P is <span class="math-tex">$\frac{x(a e)}{a^{2}}+\frac{y\left(b^{2} / a\right)}{b^{2}}=1$</span><br />
<span class="math-tex">$\Leftrightarrow \frac{x e}{a}+\frac{y}{a}=1$</span> <span class="math-tex">$\Leftrightarrow$</span> y = -xe + a<br />
<span class="math-tex">$\Rightarrow$</span> The slope of the tangent at P = - e<br />
<span class="math-tex">$\Rightarrow$</span> The slope of the normal at P = <span class="math-tex">$\frac{1}{e}$</span><br />
Equation of normal (PB') at P is<br />
<span class="math-tex">$y-\frac{b^{2}}{a}=\frac{1}{e}(x-a e)$</span><br />
The normal PB' passes through (0, -b).<br />
<span class="math-tex">$\Rightarrow-\left(b+\frac{b^{2}}{a}\right)=\frac{-1}{e}(a e)$</span><br />
<span class="math-tex">$\Leftrightarrow$</span> ab + b<sup>2</sup> = a<sup>2</sup><br />
<span class="math-tex">$\Leftrightarrow$</span> ab + a<sup>2</sup> - a<sup>2</sup>e<sup>2</sup> = a<sup>2</sup><br />
<span class="math-tex">$\Leftrightarrow$</span> ae<sup>2</sup> = b <span class="math-tex">$\Rightarrow \frac{b}{a}=e^{2}$</span><br />
<span class="math-tex">$\therefore$</span> b<sup>2</sup> = a<sup>2</sup>(1 - e<sup>2</sup>) <span class="math-tex">$\Rightarrow$</span> (e<sup>2</sup>)<sup>2</sup> = 1 - e<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> e<sup>4</sup> + e<sup>2</sup> - 1 = 0</p>
Correct Answer: D