JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 14)
The locus of the point of intersection of the lines $$\left( {\sqrt 3 } \right)kx + ky - 4\sqrt 3 = 0$$ and $$\sqrt 3 x - y - 4\left( {\sqrt 3 } \right)k = 0$$ is a conic, whose eccentricity is _________.
Answer
2
Explanation
$$\sqrt 3 kx + ky = 4\sqrt 3 $$ ........(1)
$$\sqrt 3 kx - ky = 4\sqrt 3 {k^2}$$ ....... (2)
Adding equation (1) & (2)
$$2\sqrt 3 kx = 4\sqrt 3 ({k^2} + 1)$$
$$x = 2\left( {k + {1 \over k}} \right)$$ ......... (3)
Substracting equation (1) & (2)
$$y = 2\sqrt 3 \left( {{1 \over k} - k} \right)$$ ........(4)
$$\therefore$$ $${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 4$$
$${{{x^2}} \over {16}} - {{{y^2}} \over {48}} = 1$$ (Hyperbola)
$$ \therefore $$ $${e^2} = 1 + {{48} \over {16}}$$
$$ \Rightarrow $$ $$e = 2$$
$$\sqrt 3 kx - ky = 4\sqrt 3 {k^2}$$ ....... (2)
Adding equation (1) & (2)
$$2\sqrt 3 kx = 4\sqrt 3 ({k^2} + 1)$$
$$x = 2\left( {k + {1 \over k}} \right)$$ ......... (3)
Substracting equation (1) & (2)
$$y = 2\sqrt 3 \left( {{1 \over k} - k} \right)$$ ........(4)
$$\therefore$$ $${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 4$$
$${{{x^2}} \over {16}} - {{{y^2}} \over {48}} = 1$$ (Hyperbola)
$$ \therefore $$ $${e^2} = 1 + {{48} \over {16}}$$
$$ \Rightarrow $$ $$e = 2$$
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