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  • The point of intersection of the tangents drawn to the curve   at the point where it

The point of intersection of the tangents drawn to the curve xy^{2}=y  at the point where it is met by the curve  xy=1-y is given by :

Option: 1

(2,-1)


Option: 2

(1,2)


Option: 3

(0,1)


Option: 4

None of these


Answers (1)

best_answer

Equation of curve

x y=1-y\\

\Rightarrow y=1(1+x)

The other curve is x y^2=1-y \quad \rightarrow (1)

Putting the value of  y  in equation (1)

\begin{aligned} & \Rightarrow \quad x\left\{\frac{1}{(1+x)}\right\}^2=1-\left\{\frac{1}{(1+x)}\right\} \\ & \Rightarrow \quad x\left\{\frac{1}{\left(1+x^2+2 x\right)}\right\}=1-\left\{\frac{1}{(1+x)}\right\} \\ & \Rightarrow \quad\left\{\frac{x}{\left(1+x^2+2 x\right)}\right\}=\left\{\frac{x}{(1+x)}\right\} \\ & \Rightarrow x=0 \end{aligned}

Therefore,x=0   and  y=1 .

Hence, option (3)  (0,1) is the correct answer.

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