Question:

If the focii of x216+y24=1\frac {x^2}{16}+\frac{y^2}{4}=1 and x2a2y23=1\frac {x^2}{a^2}-\frac{y^2}{3}=1 coincide, then value of aa is

Updated On: May 14, 2024
  • 11
  • 13\frac {1} {\sqrt {3}}
  • 22
  • 3\sqrt {3}
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The Correct Option is D

Solution and Explanation

Equation of conics
x216+y24=1,x2a2y29=1\frac{x^{2}}{16}+\frac{y^{2}}{4}=1,\,\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1
Equation of eccentricity of an ellipse
b2=a2(1e2)b^{2} =a^{2}\left(1-e^{2}\right)
4=16(1e2)4 =16\left(1-e^{2}\right)
e2=11/4=3/4\Rightarrow e^{2} =1-1 / 4=3 / 4
e=±34e =\pm \frac{\sqrt{3}}{4}
Focii of an ellipse =(±ae,0)=(\pm a e, 0)
=(±432,0)=(±23,0)=\left(\pm 4 \cdot \frac{\sqrt{3}}{2}, 0\right)=(\pm 2 \sqrt{3}, 0)
Given, focii of both conics are coincides.
(±23,0)=(±ae,0)\Rightarrow (\pm 2 \sqrt{3}, 0)=(\pm a e, 0)
[[\because Here (±ae,0)(\pm a e, 0) is focii of second conic.]
±ae=±23\Rightarrow \pm a e=\pm 2 \sqrt{3}
a2e2=12\Rightarrow a^{2} e^{2}=12
Equation of eccentricity of second conic (hyperbola)
b2=a2(e21)b^{2}=a^{2}\left(e^{2}-1\right)
b2=a2e2a2\Rightarrow b^{2}=a^{2} e^{2}-a^{2}
9=12a2\Rightarrow 9=12-a^{2}
a2=3\Rightarrow a^{2}=3
a=3\Rightarrow a=\sqrt{3}
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