Point of intersection of the lines -3 x - y -4 -3 m -0 and -3 m x - my -4 -3-0 describes-A- a hyperbolaB- a parabolaC- a conic of eccentricity 2D- a conic of eccentricity 1 - 2

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Point of intersection of the lines $\sqrt{3} x - y - 4 \sqrt{3} m = 0$ and $\sqrt{3} m x + m y - 4 \sqrt{3} = 0$ describes-

a hyperbola

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a parabola

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a conic of eccentricity 2

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a conic of eccentricity 1/2

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Solution

The correct options are
A
a hyperbola

C
a conic of eccentricity 2

Eliminating $m$ from given equation, we get $3 x^{2} - y^{2} = 48$ which is a hyperbola. It's eccentricity,

$e = \sqrt{1 + \frac{48}{16}} = 2$

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Point of intersection of the lines √3x−y−4√3m=0 and √3mx+my−4√3=0 describes-

The correct options are A a hyperbola C a conic of eccentricity 2 Eliminating m from given equation, we get 3x2−y2=48 which is a hyperbola. It's eccentricity, e=√1+4816=2

Point of intersection of the lines $\sqrt{3} x - y - 4 \sqrt{3} m = 0$ and $\sqrt{3} m x + m y - 4 \sqrt{3} = 0$ describes-

The correct options are
A
a hyperbola

C
a conic of eccentricity 2

Eliminating $m$ from given equation, we get $3 x^{2} - y^{2} = 48$ which is a hyperbola. It's eccentricity,

$e = \sqrt{1 + \frac{48}{16}} = 2$