Q108 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Mathematics

Tangents are drawn to the curve \(y=\sin x\) from the origin. The locus of the points of contact is

  • A
    \(x y=x+y\)
  • Correct answer
    B
    \(x^2 y^2=x^2-y^2\)
  • C
    \(x y=x-y\)
  • D
    \(x^2 y^2=x^2+y^2\)

Solution

Given, \(y=\sin x\) Differentiating w.r.t. \(x\), we get \(\frac{d y}{d x}=\cos x\) If tangent to \(y=\sin x\) meet at \((h, k)\) \(\left(\frac{d y}{d x}\right)_{(h, k)}=\cos h\) \(\therefore\) Equation of tangent is \(\cos h(x-h)=y-k\) Since, tangent is passing through (origin). \(\therefore \quad h \cos h=k\) But given curve is passing through \((h, k)\) \(\begin{aligned} k=\sin h \Rightarrow & \cos h=\sqrt{1-k^2} \\ \Rightarrow & \cos ^2 h=1-k^2=\frac{k^2}{h^2} \Rightarrow 1-\frac{k^2}{h^2}=k^2 \\ \Rightarrow & h^2-k^2=h^2 k^2 \end{aligned}\) Hence, locus of point of contact is \(x^2-y^2=x^2 y^2\)

Hence, the correct answer is option: 'B'.

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