SATHEE: Hyperbola 1 Question 9

Hyperbola 1 Question 9

10.

Consider a branch of the hyperbola

$x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0$

with vertex at the point $A$ . Let $B$ be one of the end points of its latusrectum. If $C$ is the focus of the hyperbola nearest to the point $A$ , then the area of the $△ A B C$ is

(a) $1 - \sqrt{2 / 3}$ sq unit

(b) $\sqrt{3 / 2} - 1$ sq unit

(d) $\sqrt{3 / 2} + 1$ sq unit

(2008, 3M)

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Answer:

Correct Answer: 10. (b)

Solution:

Given equation can be rewritten as focal chord

$\frac{(x - \sqrt{2})^{2}}{4} - \frac{(y + \sqrt{2})^{2}}{2} = 1$

For point $A (x, y), e = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$

$\begin{aligned} \Rightarrow & x - \sqrt{2} & = 2 \\ \Rightarrow & x & = 2 + \sqrt{2} \end{aligned}$

For point $C (x, y), x - \sqrt{2} = a e = \sqrt{6} \Rightarrow x = \sqrt{6} + \sqrt{2}$

Now, $A C = \sqrt{6} + \sqrt{2} - 2 - \sqrt{2} = \sqrt{6} - 2$

and $B C = \frac{b^{2}}{a} = \frac{2}{2} = 1$

$∴$ Area of $△ A B C = \frac{1}{2} \times (\sqrt{6} - 2) \times 1 = \sqrt{\frac{3}{2}} - 1$ sq unit

Hyperbola 1 Question 9

10.

Answer:

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Hyperbola 1 Question 9

10.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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