If the normal at - - - on the hyperbola x2-a2-y2-b2-1 meets the transverse axis at G- then AG-A-G -

The correct option is A Equation of the normal at θ is ax/sexθ+by/tanθ=a^2e^2 This normal meets the major axis i.e y = 0 at G = (ae^2 sec θ, 0) A = (a, 0), ...

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Standard XII

Mathematics

Directrix of Hyperbola

If the normal...

Question

If the normal at $‘ θ^{'}$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ meets the transverse axis at G, then AG.A’G =

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Solution

The correct option is A
Equation of the normal at $θ i s \frac{a x}{s e x θ} + \frac{b y}{t a n θ} = a^{2} e^{2}$
This normal meets the major axis i.e $y = 0 a t G = (a e^{2} s e c θ, 0) A = (a, 0), A^{'} = (- a, 0)$
$∴$ AG.A’G $= | a (e^{2} s e c θ - 1) | | a (e^{2} s e c θ + 1) | = a^{2} (e^{4} s e c^{2} θ - 1)$

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If the normal at ‘θ′ on the hyperbola x2a2−y2b2=1 meets the transverse axis at G, then AG.A’G =

The correct option is A Equation of the normal at θ is axsexθ+bytanθ=a2e2 This normal meets the major axis i.e y=0 at G =(ae2secθ,0)A=(a,0),A′=(−a,0) ∴ AG.A’G =|a(e2secθ−1)||a(e2secθ+1)|=a2(e4sec2θ−1)

If the normal at $‘ θ^{'}$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ meets the transverse axis at G, then AG.A’G =