Question

The line lx + my + n = 0 will be a normal to the hyperbola $b^2{x}^2-a^2{y}^2=a^2{b}^2$, if

• A. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$
  
• B. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2{)}^2}{n^2}$
  
• C. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n}$
• D. None of these

Answer 1

The correct option is D None of these

The equation of the normal at $(asec\varphi ,btan\varphi )$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
ax $sin\varphi +by=(a^2+b^2)tan\varphi$ ......(i)
and the equation of the line is
lx + my + n =0 …… (ii)
now, equations (i) and (ii) represent the same line, therefore
$\frac{2sin\varphi }{l}=\frac{b}{m}=\frac{(a^2+b^2)tan\varphi }{-n}=\frac{(a^2+b^2)sin\varphi }{-ncos\varphi }$
$\therefore$ $sin\varphi =\frac{bl}{am}andcos\varphi =\frac{(a^2+b^2)l}{-na}$
$\Rightarrow$ $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{-n^2}$