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The locus of the point of intersection of the tangents to the circle \mathrm{x=r \cos \theta, y=r \sin \theta} at points whose parametric angles are differ by \mathrm{\pi / 3}, is
 

Option: 1

\mathrm{x^2+y^2=4(2-\sqrt{3}) r^2}
 


Option: 2

\mathrm{3\left(x^2+y^2\right)=1}
 


Option: 3

\mathrm{x^2+y^2=(2-\sqrt{3}) r^2}
 


Option: 4

\mathrm{3\left(x^2+y^2\right)=4 r^2}


Answers (1)

best_answer

Let one of the points on the circle be \mathrm{P(r \cos \theta, r \sin \theta).}

Then the other point is \mathrm{\mathrm{Q}\{r \cos (\theta+\pi / 3), r \sin (\theta+\pi / 3)\}}.

\thereforeEquation of the tangent at \mathrm{P} is \mathrm{x \cos \theta+y \sin \theta=r,}

and the equation of the tangent at \mathrm{\mathrm{Q} \: is \: x \cos \left(\theta+\frac{\pi}{3}\right)+y \sin \left(\theta+\frac{\pi}{3}\right)=r}

\mathrm{ \Rightarrow \quad x\left[\cos \theta\left(\frac{1}{2}\right)-\sin \theta\left(\frac{\sqrt{3}}{2}\right)\right]+y\left[\sin \theta\left(\frac{1}{2}\right)+\cos \theta\left(\frac{\sqrt{3}}{2}\right)\right]=r }
\mathrm{ \Rightarrow \quad \frac{1}{2}(x \cos \theta+y \sin \theta)-\frac{\sqrt{3}}{2}(x \sin \theta-y \cos \theta)=r }
\mathrm{ \Rightarrow \quad \frac{r}{2}-\frac{\sqrt{3}}{2}(x \sin \theta-y \cos \theta)=r \quad \Rightarrow \quad(x \sin \theta-y \cos \theta)=-\frac{r}{\sqrt{3}}}
The locus of the point of intersection of these tangent can now be obtained by eliminating \theta
\therefore Squaring and adding (i) and (ii)

\mathrm{\Rightarrow \quad(x \cos \theta+y \sin \theta)^2+(x \sin \theta-y \cos \theta)^2=r^2+\left(-\frac{r}{\sqrt{3}}\right)^2 }
\mathrm{ \Rightarrow \quad 3\left(x^2+y^2\right)=4 r^2 .}

Hence option 4 is correct
 

Posted by

jitender.kumar

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