Find the locus of the point of intersection of the lines -3 x-y-4-3-0 and -3- x - y-4-3-0 for different values of -

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Latus Rectum of Hyperbola

Find the locu...

Question

Find the locus of the point of intersection of the lines $\sqrt{3} x - y - 4 \sqrt{3} λ = 0$ and $\sqrt{3} λ x + λ y - 4 \sqrt{3} = 0$ for different values of $λ$ .

3 x^{2} - y^{2} = 48

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y^{2} - 3 x^{2} = 24

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4 x^{2} - 3 y^{2} = 16

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None of these

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Solution

The correct option is A $3 x^{2} - y^{2} = 48$
Let $(h, k)$ be the point of intersection of the given lines. Then,
$\sqrt{3} h - k - 4 \sqrt{3} λ = 0$ and $\sqrt{3} λ h + λ k - 4 \sqrt{3} = 0$
$\sqrt{3} h - k = 4 \sqrt{3} λ$ and $λ (\sqrt{3} h + k) = 4 \sqrt{3}$
$(\sqrt{3} h - k) λ (\sqrt{3} h + k) = (4 \sqrt{3} λ) (4 \sqrt{3})$
$3 h^{2} - k^{2} = 48$
Hence, the locus of $(h, k)$ is $3 x^{2} - y^{2} = 48$ .

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Find the locus of the point of intersection of the lines √3x−y−4√3λ=0 and √3λx+λy−4√3=0 for different values of λ.

The correct option is A 3x2−y2=48Let (h,k) be the point of intersection of the given lines. Then,√3h−k−4√3λ=0 and √3λh+λk−4√3=0√3h−k=4√3λ and λ(√3h+k)=4√3(√3h−k)λ(√3h+k)=(4√3λ)(4√3)3h2−k2=48Hence, the locus of (h,k) is 3x2−y2=48.

Find the locus of the point of intersection of the lines $\sqrt{3} x - y - 4 \sqrt{3} λ = 0$ and $\sqrt{3} λ x + λ y - 4 \sqrt{3} = 0$ for different values of $λ$ .