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  • From the points on the circle , tangents are drawn to the hyperbola <img alt="x^2-y^2

From the points on the circle x^2+y^2=4, tangents are drawn to the hyperbola x^2-y^2=4. Then  the locus of the mid-points of the chord of contact is 

Option: 1

4\left(x^{2}+y^{2}\right)=\left(x^{2}-y^{2}\right)^{2}


Option: 2

\left(x^{2}+y^{2}\right)=\left(x^{2}-y^{2}\right)^{2}


Option: 3

\left(x^{2}+y^{2}\right)=4\left(x^{2}-y^{2}\right)^{2}


Option: 4

\left(x^{2}-y^{2}\right)=4\left(x^{2}+y^{2}\right)^{2}^{}


Answers (1)

best_answer

The equation of the chord of contact QR from the point P(2 cos \theta, 2 sin \theta) lying on circle to the hyperbola x^2-y^2=4 is 

T = 0

\\\Rightarrow \mathrm(2 \cos \theta) x-(2 \sin \theta) y=4\\\Rightarrow \mathrm( \cos \theta) x-( \sin \theta) y=2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)\\

Let the mid-point be M (h, k) to the hyperbola x^2-y^2=4

Then equation of this chord using T = S1 is

hx-ky=h^2-k^2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)

Since the Eqs (i) and (ii) are identical, so

\\\mathrm{\;\;\;\;}\frac{h}{\cos \theta}=\frac{-k}{\sin \theta}=\frac{h^{2}-k^{2}}{2}\\\Rightarrow \quad \cos \theta=\frac{2 h}{h^{2}-k^{2}} \text { and } \sin \theta=\frac{-2k}{h^{2}-k^{2}}

Squaring and adding, we get

\\ \begin{aligned} &\left(\frac{2 h}{h^{2}-k^{2}}\right)^{2}+\left(\frac{-2 k}{h^{2}-k^{2}}\right)^{2}=1 \\ \Rightarrow & 4\left(h^{2}+k^{2}\right)=\left(h^{2}-k^{2}\right)^{2} \end{aligned}

Hence, the locus of M(h, k) is a 4\left(x^{2}+y^{2}\right)=\left(x^{2}-y^{2}\right)^{2}

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Nehul

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