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Find the equation of the rectangular hyperbola which has for one of its asymptotes the line $x + 2 y - 5 = 0$ and passes through the points $(6, 0)$ and $(- 3, 0)$ .

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(x + 2 y - 5) (2 x - y + 4) = 16

(x - 2 y - 5) (2 x + y + 4) = 16

(x + 2 y + 5) (2 x - y - 4) = 16

(x - 2 y + 5) (2 x + y - 4) = 16

answer is A.

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Detailed Solution

We know, the asymptotes of a rectangular hyperbola is always perpendicular. So, for an equation of first asymptote be

ax + by + c = 0

then the line which is perpendicular to this asymptote will have the equation as

bx - ay + c_{1} = 0

where,

c

and

c_{1}

are constant terms.
Now,
Given: First asymptote equation:

x + 2 y - 5 = 0

Now, comparing with

ax + by + c = 0

, we get,

\Rightarrow a = 1

\Rightarrow b = 2

\Rightarrow c = - 5

Then, as stated above the asymptote which is perpendicular to this line will have the equation:

\Rightarrow bx - ay + c_{1} = 0

Putting the values of

a, b

we get,

\Rightarrow 2 x - y + c_{1} = 0

Now, we know, if there are two equations of asymptotes of a hyperbola

ax + by + c = 0

and

bx - ay + c_{1} = 0

given, then the equation of the hyperbola is given by

\Rightarrow (ax + by + c = 0) (bx - ay + c_{1} = 0) = k

, where

k

is a constant.
Therefore, the equation for the rectangular hyperbola is,

\Rightarrow (x + 2 y - 5) (2 x - y + c_{1}) = k

[equation I]
Now, this equation passes through

(6, 0)

and

(- 3, 0)

,
So, for

(6, 0)

\Rightarrow (x + 2 y - 5) (2 x - y + c_{1}) = k

\Rightarrow (6 + 2 \times 0 - 5) (2 \times 6 - 0 + c_{1}) = k

\Rightarrow (6 - 5) (12 + c_{1}) = k

\Rightarrow 1 \times (12 + c_{1}) = k

\Rightarrow (12 + c_{1}) = k

\Rightarrow c_{1} = k - 12

[equation II]
So, for

(- 3, 0)

\Rightarrow (x + 2 y - 5) (2 x - y + c_{1}) = k

\Rightarrow (- 3 + 2 \times 0 - 5) \{2 \times (- 3) - 0 + c_{1}\} = k

\Rightarrow (- 3 - 5) (- 6 + c_{1}) = k

\Rightarrow (- 8) (- 6 + c_{1}) = k

\Rightarrow (48 - 8 c_{1}) = k

[equation III]
Putting the value of

c_{1}

from equation II in equation III,

\Rightarrow \{48 - 8 \times (k - 12)\} = k

\Rightarrow [48 - 8 k - \{8 \times (- 12)\}] = k

\Rightarrow 48 - 8 k + 96 = k

\Rightarrow 48 + 96 = k + 8 k

\Rightarrow 144 = 9 k

\Rightarrow k = \frac{144}{9}

\Rightarrow k = 16

Now, put the value of

k

in equation II,

\Rightarrow c_{1} = k - 12

\Rightarrow c_{1} = 16 - 12

\Rightarrow c_{1} = 4

Now, putting the values of

k

and

c_{1}

in equation I,

\Rightarrow (x + 2 y - 5) (2 x - y + 4) = 16

Hence, the required equation for the rectangular hyperbola is

\Rightarrow (x + 2 y - 5) (2 x - y + 4) = 16

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