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  • The line <img alt="\mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bl%20%5Cmathrm%7Bx%7D&plus;%5Cmathrm%7Bmy%7D&plus;%5C

The line \mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0} will be a normal to the hyperbola \mathrm{b^2 x^2-a^2 y^2=a^2 b^2} if
 

Option: 1

\mathrm{\frac{\mathrm{a}^2}{l^2}+\frac{\mathrm{b}^2}{\mathrm{~m}^2}=\frac{\left(\mathrm{a}^2+\mathrm{b}^2\right)^2}{\mathrm{n}^2}}


Option: 2

\mathrm{\frac{\mathrm{a}^2}{l^2}-\frac{\mathrm{b}^2}{\mathrm{~m}^2}=\frac{\left(\mathrm{a}^2+\mathrm{b}^2\right)^2}{\mathrm{n}^2}}


Option: 3

\mathrm{\frac{\mathrm{a}^2}{l^2}-\frac{\mathrm{b}^2}{\mathrm{~m}^2}=\frac{\left(\mathrm{a}^2+\mathrm{b}^2\right)^2}{\mathrm{n}}}


Option: 4

none of these


Answers (1)

best_answer

Equation of normal at \mathrm{(a \sec \theta, b \tan \theta)} is
\mathrm{\text { ax } \cos \theta+b y \cot \theta=a^2+b^2}
Comparing it with\mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0} we get 
\mathrm{\frac{\mathrm{a} \cos \theta}{l}=\frac{\mathrm{b} \cot \theta}{\mathrm{m}}=\frac{\left(\mathrm{a}^2+\mathrm{b}^2\right)}{-\mathrm{n}}} and \mathrm{\cot \theta=\frac{m\left(a^2+b^2\right)}{-n b} \Rightarrow \sin \theta=\frac{b l}{a m}}
Thu\mathrm{\frac{\mathrm{b}^2 l^2}{\mathrm{a}^2 \mathrm{~m}^2}+\frac{l^2\left(\mathrm{a}^2+\mathrm{b}^2\right)^2}{\mathrm{a}^2 \mathrm{n}^2}=1 \text { or, } \frac{\mathrm{a}^2}{l^2}-\frac{\mathrm{b}^2}{\mathrm{~m}^2}=\frac{\left(\mathrm{a}^2+\mathrm{b}^2\right)^2}{\mathrm{n}^2}}
Hence (b) is the correct answer.

Posted by

SANGALDEEP SINGH

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