Question

Let the ellipse x^2+16y^2 =16 be inscribed in a rectangle whose sides are parallel to co-ordinate axes. If this rectangle is inscribed in another ellipse that passes through the point (16,0)then equation of the outer ellipse

Answer 1

Given :

Equation of elipse = x² +16y² = 16

It is inscribed in a rectangle,whose sides are parallel to co ordinate axis.

Rectangle is inscribed in another ellipse which passes through 16,0.

To find :

Equation of outer ellipse.

Solution :

General equation of ellipse :

$\\ \bf \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1$

(equation 1)

So,

our equation of inner ellipse will be,

$\\ \bf \: \: \frac{ {x}^{2} }{ 16 } + \frac{ {y}^{2} }{ 1 } = 1$

so,

our equation will be,

$\\ \bf \: \: \frac{ {x}^{2} }{ {4}^{2} } + \frac{ {y}^{2} }{ {1}^{2} } = 1$

So,

a² = 4², a = ±4

b² = 1², b = ±1

Means,

Co-ordinates of extreme right point on x axis = (4,0)

Co-ordinates of extreme left point on x axis = (-4,0)

Co-ordinates of highest point on y axis = (1,0)

Co-ordinates of lowest point on y axis = (-1,0).

(see figure).

So, in case of rectangle inscribing the ellipse,

its sides will get equations as

x = +4,

x = -4,

y = +1,

y = -1.

An another ellipse is inscribing the rectangle,

so its general equation is :

$\\ \bf \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1$

It is given that a point (16,0) lies on this ellipse,

so on observing we get that

it is the highest point on y axis of second ellipse.

It means in this case

b = 16

and

b² = 16²

Putting in equation 1, we get:

$\\ \bf \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {16}^{2} } = 1$

(equation 2)

We have found the equations of rectangle,

so, a point lies on outer ellipse, where two lines of equations x = +4 and y = +1 are crossing.

Co-ordinates of that point = (4,1)

so,

Putting these co-ordinates in equation 2,

we get

$\\ \bf \: \: \frac{ {4}^{2} }{ {a}^{2} } + \frac{ {1}^{2} }{ {16}^{2} } = 1$

solving this, we get :

$\\ \bf \: \: \frac{ 16 }{ {a}^{2} } + \frac{ 1 }{ {16}^{2} } = 1 \:$

$\\ \bf \: \: \frac{ (16 \times {16}^{2}) + {a}^{2} }{ {(16a)}^{2} } = 1 \:$

$\\ \bf ( {16}^{3}) + {a}^{2} = {(16a)}^{2} \: \:$

$\\ \bf ( {16}^{3}) = {(16a)}^{2} - {a}^{2} \: \:$

so,

$\bf \: 4096 = 255 {a}^{2} \\ \bf \: {a}^{2} = \frac{4096}{255}$

Putting the value of a² in equation 2,

The equation of outer ellipse :

$\\ \bf \: \: \frac{ {x}^{2} }{ { (\frac{4096}{255} )} } + \frac{ {y}^{2} }{ {16}^{2} } = 1$

So,

The equation of outer ellipse :

$\\ \bf \: \: \frac{(255 \times {x}^{2}) }{ 4096 } + \frac{ {y}^{2} }{ {16}^{2} } = 1$

or

$\\ \bf \: \: (255 \times {x}^{2}) + 16{y}^{2} = 4096$