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  • The equation  <img alt="\mathrm{\left|\sqrt{x^2+(y-1)^2}-\sqrt{x^2+(y+1)^2}\right|=K}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%7C%5Csqrt%7Bx%5E2&plus;%28y-

The equation  \mathrm{\left|\sqrt{x^2+(y-1)^2}-\sqrt{x^2+(y+1)^2}\right|=K} will represents a hyperbola, if

Option: 1

\mathrm{K} \in(0,2)


Option: 2

\mathrm{K} \in(0,1)


Option: 3

\mathrm{K} \in(1,2)


Option: 4

\mathrm{K} \in(0,2)


Answers (1)

best_answer

\mathrm{\left|\sqrt{x^2+(y-1)^2}-\sqrt{x^2+(y+1)^2}\right|=k}

Which is equivalent to \mathrm{\left|\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}\right|=\text { constant. }}

Where \mathrm{\mathrm{S}_1=(0,1), \mathrm{S}_2(0,-1)} and \mathrm{\mathrm{P} \equiv(\mathrm{x}, \mathrm{y})}

Using properties of hyperbola the above equation represents a hyperbola then 2a = k 

And \mathrm{2 \mathrm{ae}=\mathrm{S}_1 \mathrm{~S}_2=2}

Dividing \mathrm{e=\frac{2}{k}}

\mathrm{e>1}  for hyperbola, therefore \mathrm{k< 2}

Also k must be positive quantity

Hence \mathrm{k \in(0,2) \text {. }}

Posted by

shivangi.shekhar

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