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Hint: Let us take the equation of the ellipse as $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ , where a is the major axis and b is the minor axis. The eccentricity of the ellipse i.e. e is given by the formula, ${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$.

Complete step-by-step answer:

Before proceeding with the question, we must be familiar with the terms like major axis, minor axis, eccentricity and foci of the ellipse.

In an ellipse, if we have been given the foci as (0, be) and (0, -be), then from this, we get that the major axis of the ellipse is the y-axis. Length of the major axis is given by 2b and minor axis by 2a.

In this question we have been given an ellipse, with foci at (0, 2) and (0, -2), the minor axis of length 4. We have to find out through which of the given points the ellipse passes.

Let us assume the equation of ellipse as, $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.

We know that the foci of the ellipse are (0, 2) and (0, -2). So, the major axis of the given ellipse is the y-axis and we know that the minor axis of an ellipse is given as 2a.

As it is given the length of the minor axis is 4, we can write that,

2a = 4

$\Rightarrow a=2....(i)$

Foci of an ellipse are given by (0, be) and (0, -be).

Since one of the foci of the ellipse is (0, 2), we can write that,

be = 2

$\therefore e=\dfrac{2}{b}.....(ii)$

We know that eccentricity of the ellipse is given by the formula ${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$. So, we can substitute the value of a from equation (i) and e from equation (ii) in the formula and we will get,

$\Rightarrow {{\left( 2 \right)}^{2}}={{b}^{2}}\left( 1-{{\left( \dfrac{2}{b} \right)}^{2}} \right)$

$\Rightarrow 4={{b}^{2}}\left( 1-\dfrac{4}{{{b}^{2}}} \right)$

Taking LCM of terms inside bracket, we get,

$\Rightarrow 4={{b}^{2}}\left( \dfrac{{{b}^{2}}-4}{{{b}^{2}}} \right)$

$\Rightarrow 4=\left( {{b}^{2}}-4 \right)$

$\Rightarrow {{b}^{2}}=8$

Now we have got the value of ${{a}^{2}}$ as 4 and ${{b}^{2}}$ as 8.

We can substitute the value of ${{a}^{2}}$ and ${{b}^{2}}$ in the equation of ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and we will get,

$\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{8}=1$

Hence, the equation of the ellipse is $\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{8}=1$.

Now, we can substitute the value of (x, y) from the options to the equation of ellipse. If the option satisfies the equation of ellipse, then that option is the answer to the question.

Substituting the first option, $\left( 1,2\sqrt{2} \right)$ in the equation of the ellipse, we get

\[\begin{align}

& \dfrac{1}{4}+\dfrac{{{\left( 2\sqrt{2} \right)}^{2}}}{8}=1 \\

& \dfrac{1}{4}+\dfrac{8}{8}=1 \\

& \dfrac{1}{4}+1\ne 1 \\

\end{align}\]

Since it is not satisfying, we will check the second option, $\left( 2,\sqrt{2} \right)$.

\[\begin{align}

& \dfrac{{{2}^{2}}}{4}+\dfrac{{{\left( \sqrt{2} \right)}^{2}}}{8}=1 \\

& \dfrac{4}{4}+\dfrac{2}{8}=1 \\

& 1+\dfrac{1}{4}\ne 1 \\

\end{align}\]

Since it is not satisfying, we will check the third option, $\left( 2,2\sqrt{2} \right)$.

\[\begin{align}

& \dfrac{{{2}^{2}}}{4}+\dfrac{{{\left( 2\sqrt{2} \right)}^{2}}}{8}=1 \\

& \dfrac{4}{4}+\dfrac{8}{8}=1 \\

& 1+1\ne 1 \\

\end{align}\]

Since it is not satisfying, we will check the last option, $\left( \sqrt{2},2 \right)$.

$=\dfrac{{{\left( \sqrt{2} \right)}^{2}}}{4}+\dfrac{{{\left( 2 \right)}^{2}}}{8}$

$=\dfrac{2}{4}+\dfrac{4}{8}$

$=\dfrac{1}{2}+\dfrac{1}{2}$

$=1$

Since it satisfies the equation of the ellipse, we get that the ellipse passes through the point $\left( \sqrt{2},2 \right)$.

Hence, option (d) is the correct answer.

Note: Be careful about the coordinates of foci as foci will decide the major or minor axis of the ellipse. It is important to use the formula ${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$ whose major axis is y-axis, for finding the eccentricity of the ellipse. Any mistake in this formula can lead to the wrong answer. When checking each option, we can avoid the calculation by looking for an option that would give 2 and 4 on the numerator, such that we get both the terms as and it adds up to 1. The only option possible is $\left( \sqrt{2},2 \right)$.

Complete step-by-step answer:

Before proceeding with the question, we must be familiar with the terms like major axis, minor axis, eccentricity and foci of the ellipse.

In an ellipse, if we have been given the foci as (0, be) and (0, -be), then from this, we get that the major axis of the ellipse is the y-axis. Length of the major axis is given by 2b and minor axis by 2a.

In this question we have been given an ellipse, with foci at (0, 2) and (0, -2), the minor axis of length 4. We have to find out through which of the given points the ellipse passes.

Let us assume the equation of ellipse as, $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.

We know that the foci of the ellipse are (0, 2) and (0, -2). So, the major axis of the given ellipse is the y-axis and we know that the minor axis of an ellipse is given as 2a.

As it is given the length of the minor axis is 4, we can write that,

2a = 4

$\Rightarrow a=2....(i)$

Foci of an ellipse are given by (0, be) and (0, -be).

Since one of the foci of the ellipse is (0, 2), we can write that,

be = 2

$\therefore e=\dfrac{2}{b}.....(ii)$

We know that eccentricity of the ellipse is given by the formula ${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$. So, we can substitute the value of a from equation (i) and e from equation (ii) in the formula and we will get,

$\Rightarrow {{\left( 2 \right)}^{2}}={{b}^{2}}\left( 1-{{\left( \dfrac{2}{b} \right)}^{2}} \right)$

$\Rightarrow 4={{b}^{2}}\left( 1-\dfrac{4}{{{b}^{2}}} \right)$

Taking LCM of terms inside bracket, we get,

$\Rightarrow 4={{b}^{2}}\left( \dfrac{{{b}^{2}}-4}{{{b}^{2}}} \right)$

$\Rightarrow 4=\left( {{b}^{2}}-4 \right)$

$\Rightarrow {{b}^{2}}=8$

Now we have got the value of ${{a}^{2}}$ as 4 and ${{b}^{2}}$ as 8.

We can substitute the value of ${{a}^{2}}$ and ${{b}^{2}}$ in the equation of ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and we will get,

$\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{8}=1$

Hence, the equation of the ellipse is $\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{8}=1$.

Now, we can substitute the value of (x, y) from the options to the equation of ellipse. If the option satisfies the equation of ellipse, then that option is the answer to the question.

Substituting the first option, $\left( 1,2\sqrt{2} \right)$ in the equation of the ellipse, we get

\[\begin{align}

& \dfrac{1}{4}+\dfrac{{{\left( 2\sqrt{2} \right)}^{2}}}{8}=1 \\

& \dfrac{1}{4}+\dfrac{8}{8}=1 \\

& \dfrac{1}{4}+1\ne 1 \\

\end{align}\]

Since it is not satisfying, we will check the second option, $\left( 2,\sqrt{2} \right)$.

\[\begin{align}

& \dfrac{{{2}^{2}}}{4}+\dfrac{{{\left( \sqrt{2} \right)}^{2}}}{8}=1 \\

& \dfrac{4}{4}+\dfrac{2}{8}=1 \\

& 1+\dfrac{1}{4}\ne 1 \\

\end{align}\]

Since it is not satisfying, we will check the third option, $\left( 2,2\sqrt{2} \right)$.

\[\begin{align}

& \dfrac{{{2}^{2}}}{4}+\dfrac{{{\left( 2\sqrt{2} \right)}^{2}}}{8}=1 \\

& \dfrac{4}{4}+\dfrac{8}{8}=1 \\

& 1+1\ne 1 \\

\end{align}\]

Since it is not satisfying, we will check the last option, $\left( \sqrt{2},2 \right)$.

$=\dfrac{{{\left( \sqrt{2} \right)}^{2}}}{4}+\dfrac{{{\left( 2 \right)}^{2}}}{8}$

$=\dfrac{2}{4}+\dfrac{4}{8}$

$=\dfrac{1}{2}+\dfrac{1}{2}$

$=1$

Since it satisfies the equation of the ellipse, we get that the ellipse passes through the point $\left( \sqrt{2},2 \right)$.

Hence, option (d) is the correct answer.

Note: Be careful about the coordinates of foci as foci will decide the major or minor axis of the ellipse. It is important to use the formula ${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$ whose major axis is y-axis, for finding the eccentricity of the ellipse. Any mistake in this formula can lead to the wrong answer. When checking each option, we can avoid the calculation by looking for an option that would give 2 and 4 on the numerator, such that we get both the terms as and it adds up to 1. The only option possible is $\left( \sqrt{2},2 \right)$.

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