If the tangent drawn at A(2,1) to the curve x | vedclass

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(B) The given curve is: $x=1+\frac{1}{y^2}$  ...$(i)$Differentiating with respect to $x$: $1 = -\frac{2}{y^3} \cdot \frac{dy}{dx} \Rightarrow \frac{dy

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the angle between the tangents drawn at $A$ and $B$ is neither $0$ nor $\frac{\pi}{2}$

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(B) The given curve is: $x=1+\frac{1}{y^2}$  ...$(i)$Differentiating with respect to $x$: $1 = -\frac{2}{y^3} \cdot \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -\frac{y^3}{2}$.At point $A(2,1)$,the slope $m_1 = \left(\frac{dy}{dx}ight)_{(2,1)} = -\frac{1^3}{2} = -\frac{1}{2}$.The equation of the tangent at $A(2,1)$ is: $(y-1) = -\frac{1}{2}(x-2) \Rightarrow 2y - 2 = -x + 2 \Rightarrow x + 2y = 4$  ...(ii)To find point $B$,substitute $x = 4-2y$ into the curve equation $(i)$:$4-2y = 1 + \frac{1}{y^2} \Rightarrow 3-2y = \frac{1}{y^2} \Rightarrow 3y^2 - 2y^3 = 1 \Rightarrow 2y^3 - 3y^2 + 1 = 0$.Since $A(2,1)$ is on the curve,$y=1$ is a root. Factoring: $(y-1)^2(2y+1) = 0$.The roots are $y=1$ (at $A$) and $y=-\frac{1}{2}$ (at $B$).For $y = -\frac{1}{2}$,$x = 4 - 2(-\frac{1}{2}) = 5$. So,$B = (5, -\frac{1}{2})$.The slope of the tangent at $B(5, -\frac{1}{2})$ is $m_2 = -\frac{(-1/2)^3}{2} = -\frac{-1/8}{2} = \frac{1}{16}$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{-1/2 - 1/16}{1 + (-1/2)(1/16)} ight| = \left| \frac{-9/16}{1 - 1/32} ight| = \left| \frac{-9/16}{31/32} ight| = \left| -\frac{9}{16} \cdot \frac{32}{31} ight| = \frac{18}{31}$.Since $	an 	heta = \frac{18}{31} 
eq 0$ and $	an 	heta 
eq \infty$,the angle is neither $0$ nor $\frac{\pi}{2}$.

If the tangent drawn at $A(2,1)$ to the curve $x=1+\frac{1}{y^2}$ meets the curve again at $B$,then

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