Hyperbola
Grade 11

Question:

<p>The equation of the transverse and conjugate axes of a hyperbola are respectively x + 2y - 3 = 0, 2x - y + 4 = 0 and their respective lengths are&nbsp;<span class="math-tex">\(\sqrt{2}\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\frac{2}{\sqrt{3}}\)</span>. The equation of the hyperbola is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{5}\)</span>(2x - y + 4)<sup>2</sup> - <span class="math-tex">\(\frac{3}{5}\)</span>(x + 2y - 3)<sup>2</sup> = 1</p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{5}\)</span>(x + 2y - 3)<sup>2</sup> - <span class="math-tex">\(\frac{3}{5}\)</span>(2x - y + 4)<sup>2</sup> = 1</p>
<p style="display:inline">2(2x - y + 4)<sup>2</sup> - 3(x + 2y - 3)<sup>2</sup> = 1</p>
<p style="display:inline">2(x + 2y - 3)<sup>2</sup> - 3(2x - y + 4)<sup>2</sup> = 1</p>

Step-by-Step Solution

Key Concept: The equation of a hyperbola is determined by substituting the perpendicular distances from a point to its conjugate and transverse axes as X and Y respectively into the standard form X²/a² - Y²/b² = 1.
<p>Given, 2a =&nbsp;<span class="math-tex">$\sqrt{2}$</span><br /> <span class="math-tex">$\Rightarrow$</span>&nbsp;a =&nbsp;<span class="math-tex">$\frac{1}{\sqrt{2}}$</span><br /> Also, 2b =&nbsp;<span class="math-tex">$\frac{2}{\sqrt{3}}$</span><br /> <span class="math-tex">$\Rightarrow$</span>&nbsp;b =&nbsp;<span class="math-tex">$\frac{1}{\sqrt{3}}$</span><br /> If we take the two axes as the new coordinate system, and the point of intersection of the axes as the new origin, then in the new coordinate system, equation of the hyperbola will be:<br /> <span class="math-tex">$\frac{\mathbf{X}^{2}}{\mathbf{a}^{2}}-\frac{\mathbf{Y}^{2}}{\mathbf{b}^{2}}$</span>&nbsp;= 1<br /> <span class="math-tex">$\Rightarrow$</span>&nbsp;2X<sup>2</sup> - 3Y<sup>2</sup> = 1<br /> Let&nbsp;P(x, y)&nbsp;be the coordinates of a point on the hyperbola in original x-y system, then<br /> X =&nbsp;<span class="math-tex">$\frac{|2 x-y+4|}{\sqrt{5}}$</span>, Y =&nbsp;<span class="math-tex">$\frac{|\mathbf{x}+\mathbf{2 y}-\mathbf{3}|}{\sqrt{\mathbf{5}}}$</span>&nbsp;(<span class="math-tex">$\because$</span>&nbsp;X is the distance of a point on hyperbola from&nbsp;2x - y + 4 = 0 and Y is the distance of a point on hyperbola from x + 2y - 3 = 0)<br /> So, the required equation is<br /> <span class="math-tex">$\frac{2(2 x-y+4)^{2}}{5}-\frac{3(x+2 y-3)^{2}}{5}$</span>&nbsp;= 1</p>
Correct Answer: A

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