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JEE Main 202224 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2022 · 24 June · Morning Shift · Mathematics · Hyperbola

Let - 2y = μ be a tangent to the hyperbola a ^ 2 x ^ 2 - y ^ 2 = b...

Let \lambdax−2y=μbeatangenttothehyperbola\lambdax - 2y = \mu be a tangent to the hyperbola{a}^{2}{x}^{2} - {y}^{2} = {b}^{2}.Then. Then{\left(\dfrac{\lambda}{a}\right)}^{2} - {\left(\dfrac{\mu}{b}\right)}^{2} is equal to

Options

  1. A−2- 2
  2. B−4- 4
  3. C22
  4. D44

Correct answer

D. 44

Step-by-step solution

Given that \lambdax−2y=μ\lambdax - 2y = \mu is a tangent to the curve a2x2−y2=b2{a}^{2}{x}^{2} - {y}^{2} = {b}^{2}

So a2x2−(\lambdax−μ2)2=b2{a}^{2}{x}^{2} - {\left(\dfrac{\lambdax - \mu}{2}\right)}^{2} = {b}^{2}

⇒(4a2−λ2)x2+2λ\mux−μ2−4b2=0⇒ \left(4{a}^{2} - {\lambda}^{2}\right){x}^{2} + 2\lambda\mux - {\mu}^{2} - 4{b}^{2} = 0

So for the line to be tangent, the condition is discriminant =0= 0

i.e. 4λ2μ2+4(4a2−λ2)(μ2+4b2)=04{\lambda}^{2}{\mu}^{2} + 4\left(4{a}^{2} - {\lambda}^{2}\right)\left({\mu}^{2} + 4{b}^{2}\right) = 0

4λ2b2−4a2μ2=16a2b24{\lambda}^{2}{b}^{2} - 4{a}^{2}{\mu}^{2} = 16{a}^{2}{b}^{2}

Hence, λ2a2−μ2b2=4\dfrac{{\lambda}^{2}}{{a}^{2}} - \dfrac{{\mu}^{2}}{{b}^{2}} = 4

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