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  • P is a variable point on the line y = 4, tangents are drawn to the circle <img alt="\mathrm{x^2+y^2=4}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2%3D4%7D"

P is a variable point on the line y = 4, tangents are drawn to the circle \mathrm{x^2+y^2=4} from P to touch it at A and B. The parallelogram PAQB is completed. The locus of the point Q is 

\mathrm{\left(k_2+y\right)\left(x^2+y^2\right)=k_1 y^2 \text {, where } k_2-k_1=}

 

Option: 1

2


Option: 2

0


Option: 3

-2


Option: 4

4


Answers (1)

best_answer

 P lies on y = 4 and hence its co-ordinates be taken as \mathrm{\left(h_1, 4\right)},  then AB is the chord of contact with respect to circle \mathrm{x^2+y^2=4} whose equation is 

\mathrm{h_1 x+4 y=4 \ldots . \text { (i), }}Solving with circle we get

\mathrm{\begin{aligned} & \boldsymbol{x}^2+\left(\frac{4-\boldsymbol{h}_1 \boldsymbol{x}}{4}\right)^2=4 \\ & \Rightarrow \mathrm{x}^2\left(16+\mathrm{h}^2\right)-8 \mathrm{hx}-48=0 \end{aligned}}

Above gives abscissas of the points A and B 

\mathrm{\boldsymbol{x}_1+\boldsymbol{x}_2=\frac{8 \boldsymbol{h}_1}{16+\boldsymbol{h}_1^2}}

also the points A and B lie on equation (i)

\mathrm{\therefore 4\left(\boldsymbol{y}_1+\boldsymbol{y}_2\right)=8-\frac{\boldsymbol{h}_1 8 \boldsymbol{h}_1}{16+\boldsymbol{h}_1^2} \Rightarrow \boldsymbol{y}_1+\boldsymbol{y}_2=\frac{32}{16+\boldsymbol{h}_1^2}}

Now if the point Q be (h, k), then the figure PAQB being a parallelogram its diagonals bisect

\mathrm{\begin{aligned} & \therefore \boldsymbol{x}_1+\boldsymbol{x}_2=\boldsymbol{h}_1+\boldsymbol{h}=\frac{8 \boldsymbol{h}_1}{16+\boldsymbol{h}_1^2} \\ & \boldsymbol{y}_1+\boldsymbol{y}_2=4+\boldsymbol{k}=\frac{32}{16+\boldsymbol{h}_1{ }^2} \\ & \end{aligned}}---------(ii) & (iii)

Now we have to eliminate the variable between (ii) &(iii) to find the locus of Q i.e. (h, k)

Dividing (ii) by (iii) \mathrm{\begin{aligned} & \Rightarrow \frac{h_1+h}{4+k}=\frac{h_1}{4} \quad \therefore 4 h=h_1 k \\ & h_{\text {or }}=\frac{4 h}{k} \\ & \end{aligned}}

putting value of h1 in equation (iii)

\mathrm{(4+k)\left[16+\frac{16 h^2}{k^2}\right]=32}

\mathrm{\therefore } Locus is \mathrm{(4+\boldsymbol{k})\left[\frac{16 \boldsymbol{k}^2+16 \boldsymbol{h}^2}{\boldsymbol{k}^2}\right]=32 \text { orl }(4+\mathrm{y})\left(\mathrm{x}^2+y^{2})=2y^{2}\right. } 

 

 

Posted by

Rakesh

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