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Page history last edited by PBworks 18 years, 2 months ago

Learning Objectives


Problem 1 of 3


The use of any calculator is NOT allowed for this question.


A conic section is represented by the equation 4x2 - 8x - y2 + 2y - 13 = 0.


(a) Write the equation in standard form.


(b) Sketch the graph of this conic section.






4x2 - 8x - y2 + 2y - 13 = 0

4 (x2 - 2x) - y2 + 2y = 13

4 (x2 - 2x + 1) - y2 + 2y +1 = 13 + 4 + 1

4 (x - 1)2 - (y + 1)2 = 18

2 (x - 1)2 / 9 - (y + 1)2 / 18 = 1


Note: Parameter a is not equal to parameter c and they also have different signs. This tells us that the equation is a hyperbola.


1st Step : Common factor of '4' within the x's therefore we factor out the 4. We also make the equation equal to 13.

2nd Step : We 'Complete the Square' both the x and the y values to give us nice trinomals to work with. In order to 'Complete the Square' you take half of b and then square the answer to give you c for the trinomial. Also since what we do to one side of an equation we must do to another, we add both 4 and 1 to 13. We add 4 because technically we are not adding just 1 but because the 4 is outside of the bracket we multiply the 4 with the one. Thus why we add 4.

3rd Step : Afterwards we simplify the trinomial by factoring.

4th Step : Since we know the equation is hyperbola, we also know that it has to equal to 1. In order to do that we divide both sides by 18 making the equation equal to 1. On the other side we do the necessary canceling and division to simplify the equation.



From the Equation we know the centre is (1,-1). Since a is equal to 9; the square root of 9 is 3. Therefore we know that the vertices of the hyperbola are 3 units away from the centre. Since b is equal to 18 and the square root 18 is close to 4 so we know that b is approximately just a little past 4 units of the centre. From there we build out box as a dotted line. With the box we can now find the asymtopes of the hyperbola by drawing dotted lines to opposite corners of the box. From there we can built our hyperbola, as we know that the hyperbola can not touch the asymtopes so we just follow the lines.


Problem 2 of 3


The use of any calculator is NOT allowed for this question.


Determine the equation of the ellipse x2 + 4y2 − 16 = 0, after each of the following transformations:


(a) translated two units to the right.


(b) translated three units down.


(c) expanded by a factor of two along the horizontal axis.


(d) expanded by a factor of one quarter along the vertical axis.




*first lets change that equation to standard form first:


x2 + 4y2 − 16 = 0

x2 + 4y2 = 16 'by bringing the 16 to the other side of the equal sign,-16 becomes a posetive

x2/16 + 4y2/16 = 1 'dividing everything by 16

x2/16 + y2/4 = 1 '4 goes into 16, 4 times so the denominator becomes 4 instead of 16

(a) x-22/16 + y2/4 = 1 'since the question says "translated right 2 units, "-" means its going to the right


(b) x-22/16 + y-32/4 = 1 'since the question says "translated 3 units down","-" means down


(c) x-22/16(2) + y-32/4 = 1 'to expand the ellipse by 2 on the horizontal axis,multiply the "x" denominator by 2

x-22/32 + y-32/4 = 1


(d) x-22/32 + y-32/4(1/4) = 1 'to compress the ellipse vertically by 1/4,multiply the "y" denominator by 1/4

x-22/32 + y-32 = 1 '(4/1) * (1/4) = 4/4 = 1




Problem 3 of 3


The use of any calculator is NOT allowed for this question.


The equation of a circle is given by: x2 + y2 - 4x + y - 1 = 0.


(a) Find the coordinates of the centre.


(b) Find the radius.


(c) Sketch the graph.




To get useful information from this equation x2 + y2 - 4x + y - 1 = 0. We would first have to complete it by completing the square for each term.




a) To get the coordinates of the centre, you would get it right out of the equation.


b) You can also get the radius from the equation.


c)The equation can also give you all the information you need to know to draw the graph.

From the equation, we know that the center is ( 2, -1/2), and the radius is root 21/4.


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