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Question 5 for Episode 5 of Double Oh 3.14!

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- 1. Question 5 What’s the equation?
- 2. Question <ul><li>Awwwuh. Agent Double Oh 3.14 is depressed because she thinks she’s hit a dead end. But wait! What’s this letter being read by Chef PiRsquared from Dr. Ping?! </li></ul><ul><li>A parabola that has it’s vertex at (sin2 α + tan2 α , sin( )) which is located 3(cos( ) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Hint: “a” is equal to the sum of an infinite geometric sequence where the 4 th term is and the 5 th term is . </li></ul>
- 3. DO NOT MOVE ON UNTIL YOU HAVE ANSWERED THE QUESTION OR YOU NEED HELP!
- 4. Things You Should Know <ul><li>This question contains content from Sequences, Identities and Conics. </li></ul><ul><li>First let’s deal with the hint. </li></ul><ul><li>Hint: “a” is equal to the sum of an infinite geometric sequence where the 4 th term is and the 5 th term is . </li></ul>
- 5. Finding “a” <ul><li>Ok, so to find the sum of an infinite geometric sequence, we use the formula: </li></ul><ul><li>where </li></ul><ul><li>t 1 = first term </li></ul><ul><li>r= common ratio </li></ul><ul><li>Well to find the common ratio, we just divide one term by the term immediately before it. We have terms 4 and 5 so we can find the common ratio. </li></ul>
- 6. Finding “a” <ul><li>sd </li></ul>So now we know that the common ratio is 2/3. Now we need to find the first term. * When dividing fractions, multiply the two by the reciprocal of the fraction in the denominator.
- 7. Finding “a” <ul><li>To find the first term, we use the following formula: </li></ul><ul><li>t n =t 1 r n-1 </li></ul><ul><li>Where </li></ul><ul><li>t n =“n”th term </li></ul><ul><li>t 1 =first term </li></ul><ul><li>r= common ratio </li></ul><ul><li>n=rank of “n”th term </li></ul><ul><li>We know the 4 th term: 8/27. We know the common ratio: 2/3. We know the rank of the “n”th term: 4. Now we can plug it into the equation to find the 1 st term. </li></ul>
- 8. Finding “a” Divide each side by 8/27. Now we have the 1 st term. Now we can plug it in to the other equation and solve for the sum!
- 9. Finding “a” t 1 =1 r=2/3 * When dividing fractions, multiply the two by the reciprocal of the fraction in the denominator. Hooray! We found the sum of the sequence! We now also have the value of “a”. Now let’s go back to the question.
- 10. Putting in “a”. <ul><li>A parabola that has it’s vertex at (sin2 α + tan2 α , sin( )) which is located 3(cos( ) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Well now we can put “a” in where needed and simplify. </li></ul>a=3 The 9 and 6 reduce. The 4 and 2 reduce. Now let’s rewrite the question.
- 11. Finding the Vertex <ul><li>A parabola that has it’s vertex at (sin2 α + tan2 α , sin( )) which is located 3(cos( ) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Cool! Now let’s find the coordinates of the vertex! We’ll start with the x-coordinate. </li></ul><ul><li>sin2 α + tan2 α </li></ul><ul><li>This value comes out to a number which can be found using some identities. So let’s start massaging! </li></ul>
- 12. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>
- 13. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>
- 14. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1)
- 15. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1)
- 16. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1)
- 17. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1)
- 18. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1) *one of the cos α in the cos 2 α reduces with the cos α in the denominator.
- 19. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1) *one of the cos α in the cos 2 α reduces with the cos α in the denominator.
- 20. Finding the Vertex <ul><li>sin2 α + tan2 α </li></ul>*tan 2 α -1=sec 2 α therefore 1-tan 2 α =-sec 2 α (multiplied both sides by -1) *one of the cos α in the cos 2 α reduces with the cos α in the denominator. So the x coordinate of the vertex is 0.
- 21. Finding the Vertex <ul><li>A parabola that has it’s vertex at (0, sin( )) which is located 3(cos( ) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Now we’ll find the y coordinate. </li></ul><ul><li>This one is easy. </li></ul><ul><li>means that you’ve gone the full way around the circle and you are back at coordinates (1,0). Since you are looking for the sine of this angle, you are looking at the y coordinate of this angle. The y coordinate of this angle is 0 so sin (2 π )=0. </li></ul>
- 22. Finding the Distance <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3(cos( )) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Cool! Now we’ve got the vertex. The next thing we have to do is find how far away the vertex is from the focus point. </li></ul><ul><li>3(cos( )) </li></ul>
- 23. Finding the Distance <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3(cos( )) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Cool! Now we’ve got the vertex. The next thing we have to do is find how far away the vertex is from the focus point. </li></ul><ul><li>3(cos( )) </li></ul><ul><li>3(cos(2 π )) </li></ul>
- 24. Finding the Distance <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3(cos( )) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Cool! Now we’ve got the vertex. The next thing we have to do is find how far away the vertex is from the focus point. </li></ul><ul><li>3(cos( )) </li></ul><ul><li>3(cos(2 π )) </li></ul><ul><li>3(1) </li></ul>Let’s look again at the unit circle.
- 25. Finding the Distance <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3(cos( )) units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Cool! Now we’ve got the vertex. The next thing we have to do is find how far away the vertex is from the focus point. </li></ul><ul><li>3(cos( )) </li></ul><ul><li>3(cos(2 π )) </li></ul><ul><li>3(1) </li></ul>Let’s look again at the unit circle. Go 2 π radians around the circle again. Now we are looking for the cosine value or the x coordinate. In this case, it’s 1.
- 26. Finding the Distance <ul><li>A parabola that has it’s vertex at (0, 0) </li></ul><ul><li>which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Cool! Now we’ve got the vertex. The next thing we have to do is find how far away the vertex is from the focus point. </li></ul><ul><li>3(cos( )) </li></ul><ul><li>3(cos(2 π )) </li></ul><ul><li>3(1) </li></ul><ul><li>3 </li></ul>Cool! Now we know how far the vertex is from the focus point. Now let’s look at the question again.
- 27. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Well this question seems kind of easy doesn’t it? Maybe Dr. Ping has something up his sleeve. </li></ul><ul><li>But for now, let’s draw the graph! </li></ul>
- 28. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Well this question seems kind of easy doesn’t it? Maybe Dr. Ping has something up his sleeve. </li></ul><ul><li>The first thing we’ll do is mark the vertex. </li></ul>
- 29. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Well this question seems kind of easy doesn’t it? Maybe Dr. Ping has something up his sleeve. </li></ul><ul><li>We also know that the vertex is 3 units above the focus point. This means that the focus point is at (0, 0-3) or (0, -3). </li></ul>
- 30. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Well this question seems kind of easy doesn’t it? Maybe Dr. Ping has something up his sleeve. </li></ul><ul><li>Now the parabola always opens towards the focus point so it opens down. </li></ul><ul><li>This graph isn’t very accurate, but your graph should look more or less like this. </li></ul>
- 31. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Now to write the equation. We can see that this is a vertical parabola (opens up/down). </li></ul><ul><li>The standard form for the equation of a vertical parabola is: (y-k)=4p(x-h) 2 </li></ul><ul><li>(h, k) is the vertex. </li></ul><ul><li>p is the distance between the vertex and the focus point or the directrix . </li></ul><ul><li>* directrix : a line that is p units away from the vertex. </li></ul>
- 32. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Now to write the equation. We can see that this is a vertical parabola (opens up/down). </li></ul><ul><li>The standard form for the equation of a vertical parabola is: (y-k)=4p(x-h) 2 </li></ul><ul><li>(h, k) is the vertex. </li></ul><ul><li>p is the distance between the vertex and the focus point or the directrix . </li></ul><ul><li>* directrix : a line that is p units away from the vertex. The distance between any point on the parabola and the directrix is the same as the distance between the same point and the focus point. (definition of a parabola) </li></ul>
- 33. Finding the Equation <ul><li>A parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point. What is the equation of the parabola? </li></ul><ul><li>Well this question seems kind of easy doesn’t it? Maybe Dr. Ping has something up his sleeve. </li></ul><ul><li>(y-k)=4p(x-h) 2 </li></ul><ul><li>(h, k) is the vertex. </li></ul><ul><li>p is the distance between the vertex and the focus point or the directrix . </li></ul><ul><li>Now to find the equation. </li></ul><ul><li>vertex = (0,0) </li></ul><ul><li>p=3 </li></ul><ul><li>(y-0)=4(3)(x-0) 2 </li></ul><ul><li>y=12x 2 </li></ul>
- 34. Hooray! Mary found the equation! Does she have the courage to go back now?
- 35. Click on the “xD” just above this presentation to move on to the season finale!

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