Let C _ 1 be the graph of xy = 1 and the reflection of C _ 1 in the line y = 2x is C _ 2 . If the equation of C _ 2 is expressed as 12 x ^ 2 + bxy + c y ^ 2 + d = 0 , then the value of ( b + c + d ) is equal to

Correct answer: 6 Any point on the hyperbola xy = 1 is (t, (1)/(t)) Now, the image of (t, (1)/(t)) by the line mirror y = 2x is (x - t)/(2) = (y - (1)/(t))/(- 1) = - 2((2t - (1)/(t))/(5)) ⇒ 5x = (4)/(t) - 3t and 5y = 4t + (3)/(t) Now eliminating t from above two equations we get, 12 x ^ 2 - 12 y ^ 2 - 7xy + 25 = 0 ∴ b + c + d = 6

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

Let C _ 1 be the graph of xy = 1 and the reflection of C _ 1 in the...

Let

{C}_{1}

be the graph of

xy = 1

and the reflection of

{C}_{1}

in the line

y = 2x

{C}_{2} .

If the equation of

{C}_{2}

is expressed as

12{x}^{2} + bxy + c{y}^{2} + d = 0

, then the value of

( b + c + d )

is equal to

Correct answer

Step-by-step solution

Any point on the hyperbola

xy = 1

is \left(t, \dfrac{1}{t}\right)

<br />Now, the image of \left(t, \dfrac{1}{t}\right)

by the line mirror

y = 2x

\dfrac{x - t}{2} = \dfrac{y - \dfrac{1}{t}}{- 1} = - 2\left(\dfrac{2t - \dfrac{1}{t}}{5}\right)

⇒ 5x = \dfrac{4}{t} - 3t

and

5y = 4t + \dfrac{3}{t}

Now eliminating

t

from above two equations we get,

12{x}^{2} - 12{y}^{2} - 7xy + 25 = 0

∴ b + c + d = 6

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