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Consider two concentric circles C_1: x^2+y^2-1=0 and C_2: x^2+y^2-4=0. A parabola is drawn through the points, where C_1 meets the x-axis and having arbitrary tangent of C_2 as it's directrix. Then, the locus of the focus of drawn parabola is

Option: 1

\frac{4}{3} x^2-y^2=3


Option: 2

\frac{3}{4} x^2-y^2=3


Option: 3

\frac{4}{3} x^2+y^2=3

 


Option: 4

\frac{3}{4} x^2+y^2=3


Answers (1)

best_answer

(d) Clearly, the parabola should pass through (1,0) and (-1,0).

Let directrix of the parabola be x \cos \theta+y \sin \theta=2 If S(h, k) be the focus of this parabola, then distance of ( \pm 1,0) from S and from the directrix should be same.

\begin{array}{ll} \therefore & (h-1)^2+k^2=(\cos \theta-2)^2 \\ \\\text { and } & (h+1)^2+k^2=(\cos \theta+2)^2 \end{array}

On subtracting Eq. (i) from Eq. (ii), we get

                4 h=8 \cos \theta \Rightarrow \cos \theta=\frac{h}{2}

On adding Eqs. (i) and (ii), we get

                \begin{aligned} \Rightarrow & & 2\left(h^2+k^2+1\right) & =2\left(\cos ^2 \theta+4\right) \\ \\\Rightarrow & & h^2+k^2+1 & =4+\frac{h^2}{4} \\ \\\Rightarrow & & \frac{3}{4} h^2+h^2 & =3 \end{aligned}

Hence, locus of focus is \frac{3}{4} x^2+y^2=3

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