Conic Sections

4. The parabola y² = 8x is transformed by a vertical stretch, scale factor 2, and then a horizontal translation of -1 unit. Write down the equation of the resulting parabola in the form y = f(x).
(this is not an exam question)

Answer is
y = 4 √2(x+1)

What did I screw up this time?

Should be $\displaystyle (\frac{y}{2})^2=8(x+1)$ instead. Your working suggests it's a vertical stretch by a factor of 1/2.

1.a) Sketch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where a = 3 and b = 2

b) Write down (i) the length of the major axis

(ii) the length of the minor axis

c) Draw the line y = x on your sketch.

d) Find the co-ordinates of the points of intersection of the line and the ellipse.

e) Reflect the ellipse in the line y = x and write down the equation of the resulting ellipse.



I'm on d)
Somehow the x values are x = 2,3... how?

Because $\sqrt{4x^2+9y^2}\not= 2x+3y$. Take $x=y=1$ as a counterexample.

Because $\sqrt{4x^2+9y^2}\not= 2x+3y$. Take $x=y=1$ as a counterexample.

Because $\sqrt{4x^2+9y^2}\not= 2x+3y$. Take $x=y=1$ as a counterexample.

$2x^2 + 3y^2$ ?