The points of contact of the tangents drawn from the origin to the curve y=sin⁡x lie on the curve

# The points of contact of the tangents drawn from the origin to the curve $y=\mathrm{sin}x$ lie on the curve

1. A

${x}^{2}-{y}^{2}=xy$

2. B

${x}^{2}+{y}^{2}={x}^{2}{y}^{2}$

3. C

${x}^{2}-{y}^{2}={x}^{2}{y}^{2}$

4. D

none of these

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### Solution:

Let ) be a point of contact of the tangents drawn from the origin to . Then, () lies on .

…(i)

Now,

$y=\mathrm{sin}x⇒\frac{dy}{dx}=\mathrm{cos}x⇒{\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=\mathrm{cos}h$

The equation of the tangent at () is

$y-k=\left(\mathrm{cos}h\right)\left(x-h\right)$

It passes through (0, 0).

..(ii)

From (i) and (ii), we get

${k}^{2}+\frac{{k}^{2}}{{h}^{2}}=1$                                   [Squaring and adding]

$⇒{h}^{2}-{k}^{2}={k}^{2}{h}^{2}$

Hence, the locus of (h, k) is ${x}^{2}-{y}^{2}={x}^{2}{y}^{2}$.

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