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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

The equation | x^ 2 + (y - 1) ^ 2 - x^ 2 + (y + 1) ^ 2 | = K will...

The equation \left|\sqrt{x^{2} + {\left(y - 1\right)}^{2}} - \sqrt{x^{2} + {\left(y + 1\right)}^{2}}\right| = Kwillrepresentahyperbolafor−will represent a hyperbola for -

Options

  1. AK∈(0,2)K \in \left(0, 2\right)
  2. BK ∈ \left(0, 1\right)
  3. CK \in \left(1,1,\infty\right)
  4. DK \in \left(0,0,\infty\right)

Correct answer

A. K∈(0,2)K \in \left(0, 2\right)

Step-by-step solution

We have \left|\sqrt{x^{2} + {\left(y - 1\right)}^{2}} - \sqrt{x^{2} + {\left(y + 1\right)}^{2}}\right| = K<br/><br/>Whichisequivalentto∣S1P−S2P∣=Const.<br /><br />Which is equivalent to \left|{S}_{1}P - {S}_{2}P\right|= Const .

Where {S}_{1} ≡ \left(0, 1\right), {S}_{2} ≡ \left(0, -1\right).and.andP ≡ \left(x, y\right)<br/><br/>Usingpropertiesofahyperbola,theaboveequationrepresentsahyperbola,thenwehave.<br/><br/><br /><br />Using properties of a hyperbola, the above equation represents a hyperbola, then we have.<br /><br />2a = K[where[where2aisthetransverseaxisandeistheeccentricity]<br/><br/>andis the transverse axis and e is the eccentricity]<br /><br />and2ae = {S}_{1}{S}_{2} = 2<br/><br/>Dividing,wehave<br /><br />Dividing, we havee = \dfrac{2}{K}<br/><br/>Since,<br /><br />Since,e > 1forahyperbola,thereforefor a hyperbola, thereforeK < 2<br/><br/>Also,<br /><br />Also,Kmustbeapositivequantity.Hence,wehave,must be a positive quantity. Hence, we have,K \in (0,2)\left(0, 2\right) .
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