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 Options: A . &nbspa2l2+b2m2=(a2+b2)2n2 B . &nbspa2l2+b2m2=(a2−b2)2n2 C . &nbspa2l2+b2m2=(a2+b2)2n D . &nbspNone of these

The line lx + my + n = 0 will be a normal to the hyperbola b2x2−a2y2=a2b2

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

Options:

A . a2l2+b2m2=(a2+b2)2n2

B . a2l2+b2m2=(a2−b2)2n2

C . a2l2+b2m2=(a2+b2)2n

D . None of these

Answer: Option D
:
D
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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