Q.117. Ellipse

Question 117.

For the 2 unit squares given in 
adjoining drawing ,

find the maximum height of the vertical line segment AB.

For AB = 1 find the minimum area of the ellipse.

Answer 117.

If θ = angle made by the side of oblique square with the horizontal, then taking its vertex on the horizontal as origin, the coordinates of the two vertices, one of each square, through which the oblique line passes are
(1 + cosθ,1) and
[ √2 cos(45° + θ), √2 sin(45° + θ) ]
= [ (cosθ - sinθ), (cosθ + sinθ) ]
Slope of the line,

m = (1 - cosθ - sinθ) / (1 + sinθ)

Its equation as it passes through

(1 + cosθ, 1) is y - 1 = m [x - (1 + cosθ)]

=> AB = 1 - m(1 + cosθ)

= 1 - (1 + cosθ) (1 - cosθ - sinθ) /(1 + sinθ)

= (1 + sinθ - sin^2 θ + sinθ + sinθcosθ) / (1 + sinθ)

= (cos^2 θ + sinθ cosθ + 2sinθ) / (1 + sinθ)

Using Wolfram Alpha,

maxm. value of AB = 1.45656.

Infinite ellipses can circumscribe the vertices of the squares.

For AB = 1, taking the equation of the ellipse as x^2 / a^2 + y^2 / b^2 = 1,

one vertex of the square is (acosθ, bsinθ)

The area of one square is 2 * acosθ * bsinθ = 1

=> ab = 1/sin2θ

Minimum value of ab = 1 when sin2θ is maximum = 1

=> Minimum area of the circumscribing ellipse

= πab

= π sq.units.

LINK to YA!