HENCE, the point on CURVE is \[(-2,2)\]. \[\frac{dy}{DX}=2x+1\] or \[{{\left. \frac{dy}{dx} \right|}_{x=-2}}=-3\] So, the slope of the normal at \[(-2,2)\] is \[\frac{1}{3}\]. Hence, the equation of the normal is \[\frac{1}{3}(x+2)=y-2\] or \[3y=x+8\].

"> HENCE, the point on CURVE is \[(-2,2)\]. \[\frac{dy}{DX}=2x+1\] or \[{{\left. \frac{dy}{dx} \right|}_{x=-2}}=-3\] So, the slope of the normal at \[(-2,2)\] is \[\frac{1}{3}\]. Hence, the equation of the normal is \[\frac{1}{3}(x+2)=y-2\] or \[3y=x+8\].

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The equation of the normal to the curve \[y=\left| {{x}^{2}}-\left| x \right| \right|\] at\[x=-2\].

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 11 months ago

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[a] In the neighborhood of \[x=-2,y={{x}^{2}}+x\]. HENCE, the point on CURVE is \[(-2,2)\]. \[\frac{dy}{DX}=2x+1\] or \[{{\left. \frac{dy}{dx} \right|}_{x=-2}}=-3\] So, the slope of the normal at \[(-2,2)\] is \[\frac{1}{3}\]. Hence, the equation of the normal is \[\frac{1}{3}(x+2)=y-2\] or \[3y=x+8\].

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Posted on 04 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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