Question 5

Question 5

What’s the equation?

QuestionAwwwuh. 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?!

A parabola that has it’s vertex at (sin2α + tan2α, sin( ))

which is located 3(cos( ) units up from the focus point.

What is the equation of the parabola?

Hint: “a” is equal to the sum of an infinite geometric sequence where

the 4th term is and the 5th term is .

DO NOT MOVE ON UNTIL YOU HAVE ANSWERED THE QUESTION OR

YOU NEED HELP!

Things You Should Know

This question contains content from Sequences, Identities and Conics.

First let’s deal with the hint.

Hint: “a” is equal to the sum of an infinite geometric sequence where

the 4th term is and the 5th term is.

Finding “a”

Ok, so to find the sum of an infinite geometric sequence, we use the formula:

where

t1 = first term

r= common ratio

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.

Finding “a”

sd

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.

Finding “a”

To find the first term, we use the following formula:

tn=t1rn-1

Where

tn=“n”th term

t1=first term

r= common ratio

n=rank of “n”th term

We know the 4th 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 1st term.

Finding “a”

Divide each side by 8/27.

Now we have the 1st term. Now we can plug it in to the other equation and solve for the sum!

Finding “a”

t1=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.

Putting in “a”.




Well now we can put “a” in where needed and simplify.

a=3

The 9 and 6 reduce.

The 4 and 2 reduce.

Now let’s rewrite the question.

Finding the Vertex




Cool! Now let’s find the coordinates of the vertex! We’ll start with the x-coordinate.

sin2α + tan2α

This value comes out to a number which can be found using some identities. So let’s start massaging!

Finding the Vertex

sin2α + tan2α

Finding the Vertex

sin2α + tan2α

*tan2α-1=sec2α therefore 1-tan2α=-sec2α(multiplied both sides by -1)

Finding the Vertex

sin2α + tan2α


*one of the cos α in the cos2α reduces with the cos α in the denominator.

Finding the Vertex

sin2α + tan2α


*one of the cos α in the cos2α reduces with the cos α in the denominator.

So the x coordinate of the vertex is 0.

Finding the Vertex

A parabola that has it’s vertex at (0, sin( ))



Now we’ll find the y coordinate.

This one is easy.

means that you’ve gone the full wayaround the circle and you are back atcoordinates (1,0). Since you are looking for the sine of this angle, youare looking at the y coordinate of thisangle. The y coordinate of this angleis 0 so sin (2π)=0.

Finding the DistanceA parabola that has it’s vertex at (0, 0)

which is located 3(cos( )) units up from the focus point.


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.

3(cos( ))





3(cos( ))

3(cos(2π))





3(cos( ))

3(cos(2π))

3(1)

Let’s look again at the unit circle.





3(cos( ))

3(cos(2π))

3(1)

Let’s look again at the unit circle. Go 2π radians around the circle again. Now we are looking for the cosine value or the x coordinate. In this case, it’s 1.


which is located 3 units up from the focus point.



3(cos( ))

3(cos(2π))

3(1)

3 Cool! Now we know how far the vertex is from the focus point. Now let’s look at the question again.

Finding the EquationA parabola that has it’s vertex at (0, 0) which is located 3 units up from the focus point.


Well this question seems kind of easy doesn’t it? Maybe Dr. Ping has something up his sleeve.

But for now, let’s draw the graph!




The first thing we’ll do is mark the vertex.




We also know that the vertex is 3 units above the focus point. This means that the focuspoint is at (0, 0-3) or (0, -3).




Now the parabola always opens towards thefocus point so it opens down.

This graph isn’t very accurate, but your graphshould look more or less like this.



Now to write the equation. We can see that this is a vertical parabola (opens up/down).

The standard form for the equation of a vertical parabola is:(y-k)=4p(x-h)2

(h, k) is the vertex.

p is the distance between the vertex and thefocus point or the directrix.

*directrix: a line that is p units away from thevertex.



Now to write the equation. We can see that this is a vertical parabola (opens up/down).

The standard form for the equation of a vertical parabola is:(y-k)=4p(x-h)2



*directrix: a line that is p units away from thevertex. The distance between any point onthe parabola and the directrix is the sameas the distance between the same pointand the focus point. (definition of aparabola)




(y-k)=4p(x-h)2



Now to find the equation.

vertex = (0,0)

p=3

(y-0)=4(3)(x-0)2

y=12x2

Hooray! Mary found the equation! Does

she have the courage to go back

now?

Click on the “xD” just above this presentation to move on to the season finale!

Question 5

Education

Transcript of Question 5