A) Ellipses of constant eccentricity
B) Ellipses of variable eccentricity-
C) Hyperbolas of constant eccentricity
D) Hyperbolas of variable eccentricity
Correct Answer: D
Solution :
[d] Given \[\frac{xdx}{1+{{x}^{2}}}=\frac{ydy}{1+{{y}^{2}}}\] Integrating we get, \[\frac{1}{2}\log (1+{{x}^{2}})=\frac{1}{2}\log (1+{{y}^{2}})+a\] \[\Rightarrow 1+{{x}^{2}}=c(1+{{y}^{2}}),c={{e}^{2a}}\] \[{{x}^{2}}-c{{y}^{2}}=c-1\Rightarrow \frac{{{x}^{2}}}{c-1}-\frac{{{y}^{2}}}{\left( \frac{c-1}{c} \right)}=1\] ? (1) Clearly \[c>0\] as \[c={{e}^{2a}}\] Hence, the equation (i) gives a family of hyperbolas with eccentricity \[=\sqrt{\frac{c-1+\frac{c-1}{c}}{c-1}}=\sqrt{\frac{{{c}^{2}}-1}{c-1}}=\sqrt{c+1}\] if \[c\ne 1\] Thus eccentricity varies form number to member of the family as it depends on c. If c=1, it is a pair of lines \[{{x}^{2}}-{{y}^{2}}=0\]
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