The locus of the point of intersection of the lines 3 k x + k y - 4 3 = 0 and 3 x - y - 4 3 k = 0 is a conic, whose eccentricity is

The point of intersection of the lines 3 k x + k y - 4 3 = 0 and 3 x - y - 4 3 k = 0 . k = 4 3 3 x + y = 3 x - y 4 3 ⇒ 3 x 2 - y 2 = 48 ⇒ x 2 16 - y 2 48 = 1 Which is a hyperbola. b 2 = a 2 e 2 - 1 Now, 48 = 16 e 2 - 1 ⇒ e = 4 = 2

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MathematicsHyperbolaGeneralHard2 minPYQ_2021

MathematicsHardnumerical

The locus of the point of intersection of the lines $(\sqrt{3}) k x + k y - 4 \sqrt{3} = 0$ and $\sqrt{3} x - y - 4 (\sqrt{3}) k = 0$ is a conic, whose eccentricity is

Answer:

2.00

Solution:

The point of intersection of the lines $(\sqrt{3}) k x + k y - 4 \sqrt{3} = 0$ and $\sqrt{3} x - y - 4 (\sqrt{3}) k = 0$ .

$k = \frac{4 \sqrt{3}}{\sqrt{3} x + y} = \frac{\sqrt{3} x - y}{4 \sqrt{3}}$

$\Rightarrow 3 x^{2} - y^{2} = 48$

$\Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$

Which is a hyperbola.

$b^{2} = a^{2} (e^{2} - 1)$

Now, $48 = 16 (e^{2} - 1)$

$\Rightarrow e = \sqrt{4} = 2$

Stream:JEESubject:MathematicsTopic:HyperbolaSubtopic:General

⏱ 2mℹ️ Source: PYQ_2021

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