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Ellipse

Section: Mathematics

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Question : 79 of 117

Marks: +1, -0

If the point of intersections of the ellipse

\;\frac{x^{2}}{16}+\;\frac{y^{2}}{b^{2}}=1

x^{2}+y^{2}=4b, b>4

lie on the curve

y^{2}=3x^{2}

b

[16 Mar 2021 Shift 2]

12
5
6
10

Solution:

Given, equation of ellipse

\;\frac{x^{2}}{16}+\;\frac{y^{2}}{b^{2}}=1

....(i)

and equation of circle

x^{2}+y^{2}=4b

.....(ii)

From Eqs. (i) and (ii),

\;\frac{4b-y^{2}}{16}+\;\frac{y^{2}}{b^{2}}=1

\Rightarrow \;\; 4b^{3}-y^{2}b^{2}+16y^{2}=16b^{2}

\Rightarrow y^{2}=\;\frac{16b^{2}-4b^{3}}{16-b^{2}}

As,

(x, y)

lie on

y^{2}=3x^{2}

So,

y^{2}=3(4b-y^{2})

[from Eq. (ii)]

\Rightarrow \;\; 4y^{2}=12b \Rightarrow y^{2}=3b

Now,

\;\frac{16b^{2}-4b^{3}}{16-b^{2}}=3b

\Rightarrow \;\; 16b^{2}-4b^{3}=48b-3b^{3}

\Rightarrow b^{3}-16b^{2}+48b=0 \Rightarrow b(b^{2}-16b+48)=0

\Rightarrow b(b-4)(b-12)=0

As,

b>4

So,

b=12

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