Hyperbola Ques 26
- If the line $y=m x+7 \sqrt{3}$ is normal to the hyperbola $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$, then a value of $m$ is
(2019 Main, 9 April I)
(a) $\frac{3}{\sqrt{5}}$
(b) $\frac{\sqrt{15}}{2}$
(c) $\frac{2}{\sqrt{5}}$
(d) $\frac{\sqrt{5}}{2}$
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Answer:
Correct Answer: 26.(c)
Solution:
Formula:
- Given equation of hyperbola, is $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$
Since, the equation of the normals of slope $m$ to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, are given by $y=m x \mp \frac{m\left(a^{2}+b^{2}\right)}{\sqrt{a^{2}-b^{2} m^{2}}}$
$\therefore$ Equation of normals of slope $m$, to the hyperbola (i), are
$$ y=m x \pm \frac{m(24+18)}{\sqrt{24-m^{2}(18)}} $$
$\because$ Line $y=m x+7 \sqrt{3}$ is normal to hyperbola (i)
$\therefore$ On comparing with Eq. (ii), we get
$$ \begin{array}{rlrl} & & \pm \frac{m(42)}{\sqrt{24-18 m^{2}}} & =7 \sqrt{3} \\ \Rightarrow & & \pm \frac{6 m}{\sqrt{24-18 m^{2}}} & =\sqrt{3} \\ \Rightarrow & \frac{36 m^{2}}{24-18 m^{2}} & =3 \text { [squaring both sides] } \\ \Rightarrow & 12 m^{2} & =24-18 m^{2} \\ \Rightarrow & 30 m^{2} & =24 \end{array} $$
$$ \Rightarrow \quad 5 m^{2}=4 \Rightarrow m= \pm \frac{2}{\sqrt{5}} $$
BETA
