If the line x – 1 = 0 is a directrix of the hyperbola kx2 – y2 = 6, then the hyperbola passes through the point

Given hyperbola : kx26−y26=1 so eccentricity e = 1+k and directrices x=±ae⇒x=±6kk+1⇒6kk+1=1k = 2 therefore equation of hyperbola is x23−y26=1hence it passes through the point (5, −2)

If the line x – 1 = 0 is a directrix of the hyperbola kx2 – y2 = 6, then the hyperbola passes through the point

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Maths Class 12th Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the line x – 1 = 0 is a directrix of the hyperbola kx² – y² = 6, then the hyperbola passes through the point

Option 1 -

$(- 2 \sqrt[]{5}, 6)$

Option 2 -

$(- \sqrt[]{5}, 3)$

Option 3 -

$(\sqrt[]{5}, - 2)$

Option 4 -

$(2 \sqrt[]{5}, 3 \sqrt[]{6})$

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Given hyperbola : $\frac{k x^{2}}{6} - \frac{y^{2}}{6} = 1$ so eccentricity e = $\sqrt[]{1 + k}$ and directrices $x = \pm \frac{a}{e}$

$\Rightarrow x = \pm \frac{\sqrt[]{6}}{\sqrt[]{k} \sqrt[]{k + 1}} \Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k} \sqrt[]{k + 1}} = 1$

k = 2 therefore equation of hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{6} = 1$

hence it passes through the point $(\sqrt[]{5}, - 2)$

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