Question:
<p>The equation of hyperbola whose foci are (2, 4) and (-2, 4) and eccentricity is <span class="math-tex">\(\frac{4}{3}\)</span>, is</p>
<p style="display:inline">x<sup>2</sup> - (y + 2)<sup>2</sup> = 5</p>
<p style="display:inline">x<sup>2</sup> - (y - 4)<sup>2</sup> = 5</p>
<p style="display:inline"><span class="math-tex">\(\frac{x^{2}}{9}-\frac{y^{2}}{7}=\frac{1}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{4 x^{2}}{9}-\frac{4(y-4)^{2}}{7}=1\)</span></p>
Step-by-Step Solution
Key Concept: The hyperbola's equation is derived by identifying the center as the midpoint of the foci, calculating the semi-major axis 'a' from the distance between foci (2ae), and finding 'b' using the eccentricity relation $b^2 = a^2(e^2 - 1)$.
<p>Distance between foci is 4<br />
<span class="math-tex">$\Rightarrow$</span> 2ae = 4<br />
<span class="math-tex">$\Rightarrow a^{2}=\frac{9}{4} \quad \ldots\left[\because e=\frac{4}{3}\right]$</span><br />
Also, e<sup>2</sup> = 1 + <span class="math-tex">$\frac{b^{2}}{a^{2}}$</span><br />
<span class="math-tex">$\Rightarrow \frac{16}{9}-1=\frac{b^{2}}{a^{2}}$</span><br />
<span class="math-tex">$\Rightarrow b^{2}=\frac{7}{4}$</span><br />
Centre is (0, 4)<br />
Hence, equation of hyperbola is <span class="math-tex">$\frac{4 x^{2}}{9}-\frac{4(y-4)^{2}}{7}=1$</span></p>
Correct Answer: D