Question:
<p>If the line y = mx + 7<span class="math-tex">\(\sqrt3\)</span> is normal to the hyperbola <span class="math-tex">\(\frac{x^{2}}{24}-\frac{y^{2}}{18}=1\)</span>, then a value of m is</p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{15}}2\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{5}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{\sqrt{5}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{\sqrt{5}}\)</span></p>
Step-by-Step Solution
Key Concept: Apply the condition for a line $y=mx+c$ to be normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ using the slope-form formula $c = \pm \frac{m(a^{2}+b^{2})}{\sqrt{a^{2}-b^{2}m^{2}}}$.
<p>Given equation of hyperbola, is <span class="math-tex">\(\frac{x^{2}}{24}-\frac{y^{2}}{18}=1\)</span> ...(i)<br />
Since, the equation of the normals of slope m to the hyperbola <span class="math-tex">\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)</span>, are given by y = <span class="math-tex">\(m x \mp \frac{m\left(a^{2}+b^{2}\right)}{\sqrt{a^{2}-b^{2} m^{2}}}\)</span><br />
<span class="math-tex">\(\therefore\)</span> Equation of normals of slope m, to the hyperbola (i), are<br />
y = <span class="math-tex">\(m x \pm \frac{m(24+18)}{\sqrt{24-m^{2}(18)}}\)</span> ...(ii)<br />
<span class="math-tex">\(\because\)</span> Line y = mx <span class="math-tex">\(\pm\)</span> 7<span class="math-tex">\(\sqrt3\)</span> is normal to hyperbola (i)<br />
<span class="math-tex">\(\therefore\)</span> On comparing with Eq. (ii) we get<br />
<span class="math-tex">\(\pm \frac{m(42)}{\sqrt{24-18 m^{2}}}=7 \sqrt{3}\)</span> <br />
<span class="math-tex">\(\Rightarrow\)</span> <span class="math-tex">\(\pm \frac{6 m}{\sqrt{24-18 m^{2}}}=\sqrt{3}\)</span><br />
<span class="math-tex">\(\Rightarrow\)</span> <span class="math-tex">\(\frac{36 m^{2}}{24-18 m^{2}}=3\)</span> [squaring both sides]<br />
<span class="math-tex">\(\Rightarrow\)</span> 12m<sup>2</sup> =24 - 18m<br />
<span class="math-tex">\(\Rightarrow\)</span> 30m<sup>2</sup> = 24<br />
<span class="math-tex">\(\Rightarrow\)</span> 5m<sup>2</sup> = 4 <span class="math-tex">\(\Rightarrow\)</span> m = <span class="math-tex">\(\pm \frac{2}{\sqrt{5}}\)</span></p>
Correct Answer: D