Question:
<p>The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x = -4, then the equation of the normal to it at <span class="math-tex">\(\left(1, \frac{3}{2}\right)\)</span> is</p>
<p style="display:inline">2y - x = 2</p>
<p style="display:inline">4x - 2y = 1</p>
<p style="display:inline">4x + 2y = 7</p>
<p style="display:inline">x + 2y = 4</p>
Step-by-Step Solution
Key Concept: Determine the ellipse's semi-axes using the given eccentricity and directrix distance, then apply the standard point-form equation for the normal at a given point.
<p>We have, <span class="math-tex">\(e=\frac{1}{2} \text { and } \frac{a}{e}=4\)</span><br />
<span class="math-tex">\(\therefore\)</span> a = 2<br />
Now, <span class="math-tex">\(b^{2}=a^{2}\left(1-e^{2}\right)=(2)^{2}\left[1-\left(\frac{1}{2}\right)^{2}\right]=4\left(1-\frac{1}{4}\right)=3\)</span><br />
<span class="math-tex">\(\Rightarrow \quad b=\sqrt{3}\)</span><br />
<span class="math-tex">\(\therefore\)</span> Equation of the ellipse is <span class="math-tex">\(\frac{x^{2}}{(2)^{2}}+\frac{y^{2}}{(\sqrt{3})^{2}}=1\)</span><br />
<span class="math-tex">\(\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)</span><br />
Now, the equation of normal at <span class="math-tex">\(\left(1, \frac{3}{2}\right)\)</span> is<br />
<span class="math-tex">\(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}\)</span><br />
<span class="math-tex">\(\Rightarrow \quad \frac{4 x}{1}-\frac{3 y}{(3 / 2)}=4-3\)</span><br />
<span class="math-tex">\(\Rightarrow\)</span> 4x - 2y = 1</p>
Correct Answer: B