Question

The ellipse ${x^2} + 4{y^2} = 4$ is inscribed in a rectangle aligned with a coordinate axes, which is inscribed in another ellipse that passes through the point (4, 0). Then the equation of ellipse is –

$\left( a \right){x^2} + 16{y^2} = 16$

$\left( b \right){x^2} + 12{y^2} = 16$

$\left( c \right)4{x^2} + 48{y^2} = 48$

$\left( d \right)4{x^2} + 64{y^2} = 48$

Answer 1

In this particular question first draw the pictorial representation of the above problem which will give us a clear picture of what we have to find out and mark all the points as shown in the below figure then assume a standard equation of the bigger ellipse which is passing through the point (4, 0) so satisfy it so we can easily find out the value of semi-major axis so use these concepts to reach the solution of the question.

Complete step-by-step solution:

Given data:

${x^2} + 4{y^2} = 4$ is an equation of ellipse inscribed in a rectangle aligned with the coordinate axes, which is inscribed in another ellipse that passes through the point (4, 0) as shown in the below figure.

Convert the equation of the ellipse into standard equation we have,

$\Rightarrow \dfrac{{{x^2}}}{{{2^2}}} + \dfrac{{{y^2}}}{1} = 1$ Now compare this with standard equation of ellipse i.e. $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$

Therefore, a = 2, b = 1.

So the length of the semi-major axis is 2 and the length of the semi-minor axis is 1 as shown in the below figure.

Now we have to find out the equation of the ellipse passing through the point (4, 0) as shown in the above figure.

As the “y” coordinate is zero so the point (4, 0) lies on the positive side of the x-axis.

So the length of the semi-major axis of this ellipse is 4.

$\Rightarrow {a_1} = 4$

Now consider the standard equation of ellipse which is given as,

$\Rightarrow \dfrac{{{x^2}}}{{a_1^2}} + \dfrac{{{y^2}}}{{b_1^2}} = 1$........................ (1)

Now as we see that this ellipse touches the vertex of the rectangle as shown in the above figure.

So, to find out the vertex of the rectangle in the first quadrant we have,

As the given ellipse i.e. ellipse which is inscribed I the rectangle touches the rectangle at point (2, 0) on the x-axis and on point (0, 1) on the y-axis so the vertex of the rectangle in the first quadrant is (2, 1).

So the bigger ellipse is passing through the point (2, 1) so satisfy this point along with the value of ${a_1}$ we have,

$\Rightarrow \dfrac{{{2^2}}}{{{{\left( 4 \right)}^2}}} + \dfrac{{{1^2}}}{{b_1^2}} = 1$

$\Rightarrow \dfrac{4}{{16}} + \dfrac{1}{{b_1^2}} = 1$

$\Rightarrow \dfrac{1}{{b_1^2}} = 1 - \dfrac{4}{16} = \dfrac{12}{16}$

Now substitute the values in equation (1) we have,

$\Rightarrow \dfrac{{{x^2}}}{{{4^2}}} + \dfrac{{{y^2}}}{{\dfrac{12}{16}}} = 1$

$\Rightarrow \dfrac{{{x^2}}}{{16}} + \dfrac{{12{y^2}}}{16} = 1$

$\Rightarrow {x^2} + 12{y^2} = 16$

So this is the required equation of the ellipse.

Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is the standard equation of the ellipse and this ellipse is passing through the vertex of the rectangle so first find out any vertex of the rectangle as above and satisfy it into the equation of ellipse we will get the required value of semi-minor axis so that we can find out the equation of ellipse as above.