Question

The background of a printed company logo is in the shape of an ellipse with height 30 inches and width 34 inches. Find an equation for the ellipse and use it to find the vertical width, in inches, of the logo background at a distance of 6 inches horizontally from the center. (Round your answer to the nearest hundredth if necessary.)

Answer 1

The equation of an ellipse is
`(x^(2) )/(289) + (y^(2) )/(225) = 1` and the vertical width of ellipse at x = 6 inch from centre of ellipse is 28.06 inch.

Explanation:

We know that, the equation of an ellipse with horizontal major axis with centre at (0,0):

`(x^(2) )/(a^(2) ) +(y^(2) )/(b^(2) ) = 1` ------------------------------------(1)

According to the question,

`a = (34)/(2) = 17` inch⇒
`a^(2) = 17^(2) = 289`

`b = (30)/(2) = 15` inch⇒
`b^(2) = 15^(2) = 225`

After putting the value of
`a^(2)` and
`b^(2)` in the equation(1), we get

The equation of an ellipse is
`(x^(2) )/(289) + (y^(2) )/(225) = 1`

`(y^(2) )/(225) = 1- (x^(2) )/(289)`

⇒
`y^(2) = (1-(y^(2) )/(289) )*225`

⇒
`y = \sqrt{(1-(x^(2) )/(289))*225 }`-----------------------------(2)

At
`x = 6` inch from centre(0,0) of ellipse,

`y = \sqrt{( 1- (6^(2) )/(289))*225 }`

⇒
`y = 14.03` inch

Therefore, the total vertical width of ellipse at x = 6 inch from centre of ellipse =
`2*14.03` = 28.06 inch