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Hyperbola

Section: Mathematics

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Question : 66 of 98

Marks: +1, -0

The locus of the point of intersection of the lines

(\sqrt{3}) k x + k y - 4\sqrt{3}=0

\sqrt{3} x - y - 4(\sqrt{3}) k=0

is a conic, whose eccentricity is ............ .

[25 Feb 2021 Shift 1]

Your Answer:

Solution:

Given, lines are

\sqrt{3} k x + k y - 4\sqrt{3}=0

. . . (i)

\sqrt{3} x - y - 4\sqrt{3} k = 0 \;\; \ldots \;\text{ (ii) }\;

Multiply Eq. (ii)

\times k

and then adding Eqs. (i) and (ii),

\sqrt{3} k x + k y - 4\sqrt{3}=0

\;\frac{\sqrt{3} k x - k y - 4\sqrt{3} k^{2}=0}{(2\sqrt{3} x) k=4\sqrt{3}+4\sqrt{3} k^{2}}

\therefore \;\; x = \frac{4\sqrt{3}(1+k^{2})}{2\sqrt{3} k} = 2\left(k+\frac{1}{k}\right)

Subtracting Eq. (i) from Eq. (ii),

\sqrt{3} k x + k y - 4\sqrt{3}=0

\sqrt{3} k x - k y - 4\sqrt{3} k^{2}=0

\frac{- \;\; + \;\; +}{2 k y=4\sqrt{3}-4\sqrt{3} k^{2}}

y = \frac{4\sqrt{3}(1+k^{2})}{2 k} = 2\sqrt{3}\left(k+\frac{1}{k}\right)

We have,

x = 2\left(k+\frac{1}{k}\right)

and

y = 2\sqrt{3}\left(\frac{1}{k}-k\right)

\;\frac{x}{2} = \left(k+\frac{1}{k}\right)

. . . (iii)

\;\frac{y}{2\sqrt{3}} = \left(\frac{1}{k}-k\right)

. . . (iv)

Squaring and subtracting Eq. (iii) from Eq. (iv),

\;\frac{x^{2}}{4} - \;\frac{y^{2}}{12} \;\; = \left(k^{2}+\frac{1}{k^{2}}+2\right) - \left(\frac{1}{k^{2}}+k^{2}-2\right)

\;\frac{x^{2}}{4} - \;\frac{y^{2}}{12} \;\; = 4

\;\text{ or }\;\; \;\frac{x^{2}}{16} - \;\frac{y^{2}}{48} \;\; = 1

Clearly, this is a hyperbola.

\therefore \;\; e^{2} \;\; = 1 + \;\frac{b^{2}}{a^{2}}

\;\; = 1 + \;\frac{48}{16} = 1+3

\; e^{2}=4

\; e = \sqrt{4}=2

(

\because

e is positive)

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