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AP EAMCET Engineering 2016 Code C Paper

Section: Mathematics

Question : 31 of 160

Marks: +1, -0

x=3\cos\theta, y=4\sin\theta

Solution:

For an ellipse

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

General co ordinates are:

(h+a\cos\theta, k+b\sin\theta)

Here the general co ordinates are

(3\cos\theta, 4\sin\theta)

Hence

h=k=0

\& a=3, b=4

For horizontal ellipse:

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

when

b < a)

Here, it is vertical ellipse,

(\because b>a)

\therefore a=4, b=3

(i.e., we interchange normal

a

and

b

)

\Rightarrow 9=16(1-e^{2})

\Rightarrow \frac{9}{16}=1-e^{2}

\Rightarrow e^{2}=1-\frac{9}{16}=\frac{7}{16}

\Rightarrow e=\frac{\sqrt{7}}{4}

The two focii lie on (O,I as) :Distance between them:

=2ae

=2\times 4\times \frac{\sqrt{7}}{4}=2\sqrt{7}

\therefore 1

. is the correct option

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