Question:
<p>If a hyperbola passes through the point P(10, 16) and it has vertices at (<span class="math-tex">\(\pm\)</span>6, 0), then the equation of the normal to it at P is:</p>
<p style="display:inline">x + 3y = 58</p>
<p style="display:inline">x + 2y = 42</p>
<p style="display:inline">3x + 4y = 94</p>
<p style="display:inline">2x + 5y = 100</p>
Step-by-Step Solution
Key Concept: Determine the hyperbola parameters $a$ and $b$ from the vertices and point $P$, then apply the standard normal equation $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.
<p>Let the hyperbola is <span class="math-tex">$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$</span><br />
If a hyperbola passes through vertices at (<span class="math-tex">$\pm$</span>6, 0), then<br />
<span class="math-tex">$\therefore$</span> a = 6<br />
As hyperbola passes through the point P(10, 16)<br />
<span class="math-tex">$\therefore \frac{100}{36}-\frac{256}{b^{2}}=1$</span> <span class="math-tex">$\Rightarrow$</span> b<sup>2</sup> = 144<br />
<span class="math-tex">$\therefore$</span> Required hyperbola is <span class="math-tex">$\frac{x^{2}}{36}-\frac{v^{2}}{144}=1$</span><br />
Equation of normal is <span class="math-tex">$\frac{36 x}{10}+\frac{144 y}{16}$</span> = 36 + 144<br />
<span class="math-tex">$\therefore$</span> At P(10, 16) normal is<br />
<span class="math-tex">$\frac{36 x}{10}+\frac{144 y}{16}$</span> = 36 + 144<br />
<span class="math-tex">$\therefore$</span> 2x + 5y = 100</p>
Correct Answer: D