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 16.  Find the equation for the ellipse that satisfies the given conditions:

     Length of minor axis 16, foci (0, ± 6).

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Given, In an ellipse, 

 Length of minor axis 16, foci (0, ± 6).

Here, the focus of the ellipse is on the  Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

 \frac{x^2}{b^2}+\frac{y^2}{a^2}=1

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get 

2b=16\Rightarrow b=8 and c=6

Now, As we know the relation,

a^2=b^2+c^2

a=\sqrt{b^2+c^2}

a=\sqrt{8^2+6^2}

a=\sqrt{64+36}

a=\sqrt{100}

a=10

Hence, The Equation of the ellipse will be :

\frac{x^2}{8^2}+\frac{y^2}{10^3}=1

Which is 

\frac{x^2}{64}+\frac{y^2}{100}=1.

Posted by

Pankaj Sanodiya

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