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

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None of these

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Solution

The correct option is D None of these
The equation of the normal at $(a s e c ϕ, b t a n ϕ)$ to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is
ax $s i n ϕ + b y = (a^{2} + b^{2}) t a n ϕ$ ......(i)
and the equation of the line is
lx + my + n =0 …… (ii)
now, equations (i) and (ii) represent the same line, therefore
$\frac{2 s i n ϕ}{l} = \frac{b}{m} = \frac{(a^{2} + b^{2}) t a n ϕ}{- n} = \frac{(a^{2} + b^{2}) s i n ϕ}{- n c o s ϕ}$
$∴$ $s i n ϕ = \frac{b l}{a m} a n d c o s ϕ = \frac{(a^{2} + b^{2}) l}{- n a}$
$\Rightarrow$ $\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{(a^{2} + b^{2})^{2}}{- n^{2}}$

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