Question

An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points?

• A. (1,2√2)
• B. (2,√2)
• C. (2,2√2)
• D. (√2,2)

Hint:

The eccentricity of the ellipse i.e. e is calculated by the formula, a² = b²(1 - e²).

Answer 1

The Correct Option is D

In the equation of the ellipse as \(\frac{x^2}{4}​+\frac{y^2}{8}​=1\), a denotes the major axis and b denotes the minor axis. The eccentricity of the ellipse i.e. e is calculated by the formula, a² = b²(1 - e²).

Complete step-by-step answer:

In an ellipse, if we have been given the foci as (0, b_e) and (0, -b_e), then, we get that the major axis of the ellipse is the y-axis. 

• The length of the major axis is given by 2b and the minor axis by 2a.
• Here, we have an ellipse, with foci at (0, 2) and (0, -2) and a minor axis of length 4.

To Find - Points the ellipse passes.

Let us assume the equation of ellipse as \(\frac{x^2}{a^2}​+\frac{y^2}{b^2}​=1\).

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,

\(2a=4\)

\(⇒a=2\)  …....(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

\(∴e=\frac{2}{b}\)  .......(ii)

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

\(⇒(2)^2=b^2(1−(\frac{2}{b})^2)\)

\(⇒4=b^2(1−\frac{4}{b^2})\)

Taking LCM of terms inside the bracket, we get,

\(⇒4=b^2(\frac{b^2−4}{b^2})\)

\(⇒4=(b^2−4)\)

\(⇒b^2=8\)

Now we have got the value of a² as 4 and b² as 8.

We can substitute the value of a² and b² in the equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) to get, \(\frac{x^2}{4}+\frac{y^2}{8}=1\)

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

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

Substituting the first option, \((1,2\sqrt2)\) in the equation of the ellipse, we get

\(\frac{1}{4}+\frac{(2\sqrt{2})^2}{8}=1\)

\(\frac{1}{4}+\frac{8}{8}=1\)

\(\frac{1}{4}+1≠1\)

Since it is not satisfying, we will check the second option, \((2,\sqrt2)\).

\(\frac{2^2}{4}+\frac{(\sqrt2)^2}{8}=1\)

\(\frac{4}{4}+\frac{2}{8}=1\)

\(1+\frac{1}{4}≠1\)

Since it is not satisfying, we will check the third option, \((2,2\sqrt2)\).

\(\frac{2^2}{4}+\frac{(2\sqrt2)^2}{8}=1\)

\(\frac{4}{4}+\frac{8}{8}=1\)

\(1+1≠1\)

Since it is not satisfying, we will check the last option, \((\sqrt2,2)\).

=\((\frac{\sqrt2}{4})+\frac{(2)^2}{8}\)

\(=\frac{2}{4}+\frac{4}{8}\)

\(=\frac{1}{2}+\frac{1}{2}\)

=1

Since it satisfies the equation of the ellipse, we get that the ellipse passes through the point \((\sqrt2,2)\).

Hence, option (d) is the correct answer.

Answer 2

Given that \(be = 2\) and \(a = 2\) 
(here a < b) 
\(\because \, \, a^2 = b^2 (1-e^2)\)
\(\therefore \, \, \, b^2 = 8\)
\(\therefore \,\) equation of ellipse \(\frac{x^2}{4} + \frac{y^2}{8} = 1\)

Option D) satisfies the following equation.

An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through (√2,2) points.

Answer 3

Option D is correct i.e., \((\sqrt{2},2)\)

Let the equation of ellipse be

\(\frac{x_2}{a_2}​+\frac{y_2}{b_2}​=1\)  ……....(i)

Since, foci are at (0,2) and (0,−2), major axis is along the Y-axis.

so, \(be=2\)  ……....(ii)

and 2a = length of minor axis = 4 (given)

\(⇒a=2\)

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

\((\frac{2}{b})^2=1-\frac{4}{b^2}\)

\(\frac{8}{b^2}=1\)

\(⇒b^2=8\)

Thus, the equation of a required ellipse is

\(\frac{x^2}{4}​+\frac{y^2}{8}​=1\)

Now, from the option the ellipse \(\frac{x^2}{4}​+\frac{y^2}{8}​=1\) passes through points \((\sqrt{2},2)\)