SATHEE: Hyperbola 2 Question 2

Hyperbola 2 Question 2

2. 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) $\frac{\sqrt{15}}{2}$

(d) $\frac{\sqrt{5}}{2}$

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Answer:

Correct Answer: 2. (c)

Solution:

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 $y = m x \mp \frac{m (a^{2} + b^{2})}{\sqrt{a^{2} - b^{2} m^{2}}}$

$∴$ Equation of normals of slope $m$ , to the hyperbola (i), are

$y = m x \pm \frac{m (24 + 18)}{\sqrt{24 - m^{2} (18)}}$

$∵$ Line $y = m x + 7 \sqrt{3}$ is normal to hyperbola (i)

$∴$ On comparing with Eq. (ii), we get

$\begin{aligned} \pm \frac{m (42)}{\sqrt{24 - 18 m^{2}}} & = 7 \sqrt{3} \\ \Rightarrow & \pm \frac{6 m}{\sqrt{24 - 18 m^{2}}} & = \sqrt{3} \\ \Rightarrow & \frac{36 m^{2}}{24 - 18 m^{2}} & = 3 [squaring both sides] \\ \Rightarrow & 12 m^{2} & = 24 - 18 m^{2} \\ \Rightarrow & 30 m^{2} & = 24 \end{aligned}$

$\Rightarrow 5 m^{2} = 4 \Rightarrow m = \pm \frac{2}{\sqrt{5}}$

Hyperbola 2 Question 2

2. 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

Answer:

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Hyperbola 2 Question 2

2. If the line y=mx+73 is normal to the hyperbola x224−y218=1, then a value of m is

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2. 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