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QJEEM21Ma6bff: JEE Main 2021 Mathematics - Class 11 JEEM21Ma6bff Solution: The locus of the point of intersection of the lines and is a…

The locus of the point of intersection of the lines (3)kx+ky−43=0\left( {\sqrt 3 } \right)kx + ky - 4\sqrt 3 = 0 and 3x−y−4(3)k=0\sqrt 3 x - y - 4\left( {\sqrt 3 } \right)k = 0 is a conic, whose eccentricity is _________.

Step-by-Step Verified Solution

3kx+ky=43\sqrt 3 kx + ky = 4\sqrt 3 ........(1)

3kx−ky=43k2\sqrt 3 kx - ky = 4\sqrt 3 {k^2} ....... (2)

Adding equation (1) & (2)

23kx=43(k2+1)2\sqrt 3 kx = 4\sqrt 3 ({k^2} + 1)

x=2(k+1k)x = 2\left( {k + {1 \over k}} \right) ......... (3)

Substracting equation (1) & (2)

y=23(1k−k)y = 2\sqrt 3 \left( {{1 \over k} - k} \right) ........(4)

∴\therefore x24−y212=4{{{x^2}} \over 4} - {{{y^2}} \over {12}} = 4

x216−y248=1{{{x^2}} \over {16}} - {{{y^2}} \over {48}} = 1 (Hyperbola)

∴\therefore e2=1+4816{e^2} = 1 + {{48} \over {16}}

⇒\Rightarrow e=2e = 2

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