# Thread: Parabolas

1. ## Parabolas

Hi

Write down the cartesian equation of the tangent + normal to the following parabolas at the points indicated.

(a) x = cos 2t, y = cos t at the point where t = PI/4

(b) x = t + 2, y = t^2/4 + 1 at the vertex.

For (a), I keep on getting 2x - 4sqrt(2) + 4 = 0 and 4x + sqrt(2)y -1 = 0 for the tangent and the normal respectively, which is wrong.

For (b), I have no clue how to do it.

Could someone please help me?

2. Originally Posted by xwrathbringerx For (a), I keep on getting 2x - 4sqrt(2)y + 4 = 0 and 4x + sqrt(2)y -1 = 0 for the tangent and the normal respectively, which is wrong.
I also the same equations as you did. Why are they wrong?

3. Your answer was correct (except for the missing "y" ). Where did you get stuck at (b)? Finding the vertex?!

4. ## Thanx - that must mean the textbook answer's wrong Yeah, for (b), I have no clue how to find the vertex.

5. $\displaystyle y=\frac{1}{4}t^2+1,x-2=t$ $\displaystyle \Rightarrow y=\frac{1}{4}(x-2)^2+1$ and you can see that this parabola is in the vertex form: $\displaystyle y=a(x-x_v)^2+y_v$
$\displaystyle (\Rightarrow y\geqslant y_v,\forall x\text{ and }\exists x=x_v\text{ so that }y=y_v\Rightarrow (x_v,y_v)\text{ is the vertex})$
So you have $\displaystyle (x_v,y_v)=(2,1)$.

#### Search Tags

parabolas 