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The locus of all the points on the curve \(y^2=4 a\left(x+a \sin \frac{x}{a}\right)\) at which the tangent is parallel to \(x\)-axis is:
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The correct answer is:
\(\mathrm{y}^2=4 \mathrm{ax}\)
We have
\(y^2=4 a\left(x+a \sin \frac{x}{a}\right)\) ...(i)
Differentiating w.r. \(\mathrm{tx}\),
\(2 y \frac{d y}{d x}=4 a\left[1+\cos \frac{x}{a}\right]\)
Since, the tangent is parallel to \(x-\) axis, \(\frac{d y}{d x}=0\) This gives
\(4 a\left(1+\cos \frac{x}{a}\right)=0 \Rightarrow \cos \frac{x}{a}=-1 \Rightarrow \sin \frac{x}{a}=0\)
\(\therefore\) From (i) \(y^2=4 a(x+0) \Rightarrow y^2=4 a x\), which is the required locus.
\(y^2=4 a\left(x+a \sin \frac{x}{a}\right)\) ...(i)
Differentiating w.r. \(\mathrm{tx}\),
\(2 y \frac{d y}{d x}=4 a\left[1+\cos \frac{x}{a}\right]\)
Since, the tangent is parallel to \(x-\) axis, \(\frac{d y}{d x}=0\) This gives
\(4 a\left(1+\cos \frac{x}{a}\right)=0 \Rightarrow \cos \frac{x}{a}=-1 \Rightarrow \sin \frac{x}{a}=0\)
\(\therefore\) From (i) \(y^2=4 a(x+0) \Rightarrow y^2=4 a x\), which is the required locus.
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