Hyperbola
Grade 11

Question:

<p>The length of the transverse axis of a hyperbola is 7 and it passes through the point (5, 2). The equation of the hyperbola is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{4}{49} x^{2}-\frac{196}{51} y^{2}\)</span> = 1</p>
<p style="display:inline"><span class="math-tex">\(\frac{4}{49} x-\frac{196}{51} y\)</span> = 1</p>
<p style="display:inline"><span class="math-tex">\(\frac{49}{4} x^{2}-\frac{51}{196} y^{2}\)</span> = 1 </p>
<p style="display:inline"><span class="math-tex">\(\frac{4}{49} x^{2}-\frac{51}{196} y^{2}\)</span> = 1    </p>

Step-by-Step Solution

Key Concept: Identify the parameter 'a' from the transverse axis length (2a) and substitute the given point into the standard hyperbola equation to verify the result.
<p>Trick 2a = 7 or a =&nbsp;<span class="math-tex">$\frac 72$</span><br /> Also (5, 2) satisfies it, so<br /> <span class="math-tex">$\frac{4}{49}(25)-\frac{51}{196}(4)$</span>&nbsp;= 1 and a<sup>2</sup>&nbsp;=&nbsp;<span class="math-tex">$\frac{49}{4}$</span><br /> <span class="math-tex">$\Rightarrow$</span>&nbsp;a =&nbsp;<span class="math-tex">$\frac 72$</span></p>
Correct Answer: D

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