Graph of conic x^2-(y-1)^2=1 has one tangent line with positive slope that passes through origin. po

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10-2. Parabola, Ellipse, Hyperbola

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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 the tangency being $(a, b)$ then find the value of $\sin ^{-1}\left(\frac{a}{b}\right)$

$\frac{{5\pi }}{{12}}$

$\frac{\pi }{6}$

$\frac{\pi }{3}$

$\frac{\pi }{4}$

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 $)
$\sin ^{-1}\left(\frac{a}{b}\right) = \frac{\pi }{4}$

Standard 11

Mathematics

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(JEE MAIN-2022)

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normal

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 the tangency being $(a, b)$ then find the value of $\sin ^{-1}\left(\frac{a}{b}\right)$

Solution

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