Question:
<p>The equations of tangent and normal at point (3, -2) of ellipse 4x<sup>2</sup> + 9y<sup>2</sup> = 36 are</p>
<p style="display:inline"><span class="math-tex">\(\frac{x}{3}-\frac{y}{2}=1, \frac{x}{2}+\frac{y}{3}=\frac{5}{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{x}{3}+\frac{y}{2}=1, \frac{x}{2}-\frac{y}{3}=\frac{5}{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{x}{2}-\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=\frac{5}{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}-\frac{y}{2}=\frac{5}{6}\)</span></p>
Step-by-Step Solution
Key Concept: Apply the point form substitution $x^2 \to xx_1$ and $y^2 \to yy_1$ to find the tangent equation and then use the perpendicularity condition to derive the normal equation.
<p>Given, equation of ellipse is 4x<sup>2</sup> + 9y<sup>2</sup> = 36<br />
i.e., <span class="math-tex">\(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)</span><br />
The equation of the tangent at point (3, -2) is<br />
<span class="math-tex">\(\frac{(3) x}{9}+\frac{(-2) y}{4}=1\)</span><br />
<span class="math-tex">\(\Rightarrow \frac{x}{3}-\frac{y}{2}=1\)</span><br />
Equation of normal is <span class="math-tex">\(\frac{x}{2}+\frac{y}{3}=k\)</span> and it passes through point (3,-2)<br />
<span class="math-tex">\(\Rightarrow \frac{3}{2}-\frac{2}{3}=k \Rightarrow k=\frac{5}{6} \Rightarrow\)</span> Normal is given by<br />
<span class="math-tex">\(\frac{x}{2}+\frac{y}{3}=\frac{5}{6}\)</span></p>
Correct Answer: A