Question:
<p>Find the equation of axis of the given hyperbola <span class="math-tex">\(\frac{x^{2}}{3}-\frac{y^{2}}{2}=1\)</span> which is equally inclined to the axes?</p>
<p style="display:inline">y = x + 2</p>
<p style="display:inline">y = x - 1</p>
<p style="display:inline">y = x - 2</p>
<p style="display:inline">y = x + 1</p>
Step-by-Step Solution
Key Concept: For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, tangents with slope m = 1 (equally inclined to axes) are given by $y = x \pm \sqrt{a^2m^2 - b^2} = x \pm \sqrt{a^2 - b^2}$. With $a^2 = 3, b^2 = 2$, this yields $y = x \pm 1$.
<p>We have <span class="math-tex">$\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$</span><br />
Since, tangent is equally inclined to the axes i.e., tan<span class="math-tex">$\theta$</span> = 1 = m.<br />
Equation of tangent in slope form is<br />
y = mx + <span class="math-tex">$\sqrt{a^{2} m^{2}-b^{2}}$</span><br />
Here, a<sup>2</sup> = 3, b<sup>2</sup> = 2<br />
<span class="math-tex">$\Rightarrow$</span> y = 1.. x + <span class="math-tex">$\sqrt{3 \times(1)^{2}-2}$</span><br />
<span class="math-tex">$\Rightarrow$</span> y = x + 1</p>
Correct Answer: D