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The graph of the conic $ x^2 - (y - 1)^2 = 1$ has one tangent line with positive slope that passes through the origin. the point of tangency being $(a, b). $ Then Eccentricity of the conic is
$\frac{4}{3}$
$\sqrt 3 $
$2$
$none$
Solution
differentiate the curve
$2x – 2(y – 1)\frac{{dy}}{{dx}}$ $ = 0 $
${\left. {\frac{{dy}}{{dx}}} \right]_{a,\,b}} = \frac{a}{{b – 1}}$ $= $ $(m_{OP} = )$
$a^2 = b^2 – b….(1)$
Also $(a, b)$ satisfy the curve
$a^2 – (b – 1)^2 = 1$
$a^2 – (b^2 – 2b + 1) = 1$
$a^2 – b^2 + 2b = 2$
$- b + 2b = 2$ $\Rightarrow$ $b = 2$
$a =\sqrt 2 $ ($a \ne – \sqrt 2 $)
$\sin ^{-1}\left(\frac{a}{b}\right) = \frac{\pi }{4}$
Length of latus rectum = $\frac{{2{b^2}}}{a}$ = $2a $ = distance between the vertices $= 2$
Curve is a rectangular hyperbola $ \Rightarrow$ $e$ =$\sqrt 2 $
