Axes are coordinate axes- A and B are the ends of major axis and minor axis respectively- Area of OAB is 16 sq units- e -3-2 then equation of the ellipse isA- x2-32-y2-8-1B- x2-64-y2-16-1C- x2-64-y2-8-1D- x2-64-y2-32-1

The correct option is B e = √(3)/2 1/2*a*b=16 ⇒ab=32 b^2=a^2(1 e^2) b^2=a^2(1 3/4)=a^2(1/4)⇒4b^2=a^2 2b.b=32⇒ b^2=16 a=2b,a=8 x^2/64+y^2/16=1 ...

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Standard XII

Mathematics

Principal Axes of Ellipse

Axes are coor...

Question

Axes are coordinate axes. A and B are the ends of major axis and minor axis respectively.

Area of $\vartriangle$OAB is 16 sq units, e = $\frac{\sqrt{3}}{2}$ then equation of the ellipse is

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Solution

The correct option is B

e = $\frac{\sqrt{3}}{2}$

$\frac{1}{2}$ ab=16

$\Rightarrow$ ab=32

$b^{2}$ = $a^{2} (1 - e^{2})$

$b^{2}$ = $a^{2} (1 - \frac{3}{4})$ = $a^{2} (\frac{1}{4}) \Rightarrow 4 b^{2}$ = $a^{2}$

2b.b=32 $\Rightarrow b^{2}$ =16

a=2b,a=8

$\frac{x^{2}}{64} + \frac{y^{2}}{16}$ =1

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Axes are coordinate axes. A and B are the ends of major axis and minor axis respectively. Area of \(\vartriangle\)OAB is 16 sq units, e = \(\frac{\sqrt{3}}{2}\) then equation of the ellipse is

The correct option is B e = √32 12*a*b=16 ⇒ab=32 b2=a2(1−e2) b2=a2(1−34)=a2(14)⇒4b2=a2 2b.b=32⇒b2=16 a=2b,a=8 x264+y216=1

Axes are coordinate axes. A and B are the ends of major axis and minor axis respectively.

Area of \(\vartriangle\)OAB is 16 sq units, e = \(\frac{\sqrt{3}}{2}\) then equation of the ellipse is

The correct option is B

e = $\frac{\sqrt{3}}{2}$

$\frac{1}{2}$ ab=16

$\Rightarrow$ ab=32

$b^{2}$ = $a^{2} (1 - e^{2})$

$b^{2}$ = $a^{2} (1 - \frac{3}{4})$ = $a^{2} (\frac{1}{4}) \Rightarrow 4 b^{2}$ = $a^{2}$

2b.b=32 $\Rightarrow b^{2}$ =16

a=2b,a=8

$\frac{x^{2}}{64} + \frac{y^{2}}{16}$ =1