Conics foci (1 Viewer)

Aysce · Feb 6, 2012

Q. {(x+1)^2}/8 + {(y-3)^2}/4 = 1 Find the foci and directrices. I dont understand how they got the foci.

Manroop · Feb 6, 2012

whats the answer

Carrotsticks · Feb 6, 2012

Aysce said:
Q. {(x+1)^2}/8 + {(y-3)^2}/4 = 1 Find the foci and directrices. I dont understand how they got the foci.

This is an ellipse that has been shifted to the left by 1 unit, and up 3 units.

We notice that this has the same focal length as the exact same ellipse, but centred at the origin.

So what I'll do is shift that ellipse back to the origin, find the foci, then shift the ellipse back to where it was originally (which will 'carry' the foci along with it).

Carrotsticks · Feb 6, 2012

Here's some working out if you're having trouble:

$\\ $Consider the ellipse $ \frac{x^2}{8} + \frac{y^2}{4} = 1 \\\\ $We can easily find that the foci are (2,0) and (-2,0) $ \\\\ $We note that the new ellipse is the exact same one just then, but shifted to the left 1 unit, and up 3 units$. \\\\ $Hence, we will apply those transformations to the coordinates of the foci that we just found:$ \\\\ (2,0) \Rightarrow (1,3) \\\\ (-2,0) \Rightarrow (-3,3)$

Oh, and the directrices are found the same way. However this time, we need not shift it 'up' by 3 units, since the directrices are vertical lines anyway, so shifting them up or down will make no difference.

SpiralFlex · Feb 6, 2012

Drawing a pretty diagram may help.

Our new centre of the ellipse was originally $(0,0),$ now it's

$(-1,3)$

For eccentricity,

$4=8(1-e^2)$

$\therefore e =\frac{\sqrt{2}}{2}$

The coordinates of our ORIGINAL focii were $(\pm ae,0)$

$(2\sqrt{2}*\frac{2}{2},0)=(\pm 2,0)$

Since there was a shift of one units to the left, and three units up.

$(1 \pm 2,3)$

$(1,3), (-3,3)$

Our original directrices were $x=\pm \frac{a}{e}$

$x=\pm 4$

Shifting this left, since this only affects the abssicas not the ordinates.

$x= -1\pm 4$

$\therefore x=3, -5$

Carrotsticks · Feb 6, 2012

Spiral, you could have found the coordinates of the foci using the formula:

$ae=\sqrt{a^{2} - b^{2}}$

This automatically eliminates any need to find the eccentricity.

SpiralFlex · Feb 6, 2012

Carrotsticks said:
Spiral, you could have found the coordinates of the foci using the formula:

$ae=\sqrt{a^{2} - b^{2}}$

This automatically eliminates any need to find the eccentricity.

The source Ace is using requires him to use the eccentricity method for practice.

Aysce · Feb 6, 2012

Thank you guys, so much! Do you guys mind helping me with another question?

Carrotsticks · Feb 6, 2012

Aysce said:
Thank you guys, so much! Do you guys mind helping me with another question?

Not at all.

Aysce · Feb 6, 2012

<3

4b. Write the equation of the locus of a point P that moves such that its distance from (2,1) is half as its distance from the lie x+y-2=0.

Carrotsticks · Feb 6, 2012

Aysce said:
<3

4b. Write the equation of the locus of a point P that moves such that its distance from (2,1) is half as its distance from the lie x+y-2=0.

Let P be defined by the point (x,y), and define the point Q(2,1). Let the perpendicular distance between P and the line x+y-2=0 be l.

$\\ PQ = \frac{1}{2} PL \\\\ \sqrt{(x-2)^{2} + (y-1)^2} = \frac{1}{2} \times \begin{vmatrix} \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2 }} \end{vmatrix} = \frac{1}{2} \times \begin{vmatrix} \frac{x + y - 2}{\sqrt{2 }} \end{vmatrix} \\\\ $Square both sides and re-arrange a bit:$ \\\\ (x-2)^{2} + (y-1)^2 = \frac{(x+y+2)^{2}}{8}$

This is a bit messy, but the equation should be a rotated ellipse.

You can leave it like this, or you can expand fully and add/subtract like terms etc.

Conics foci (1 Viewer)

Aysce

Well-Known Member

Manroop

Member

Carrotsticks

Retired

Carrotsticks

Retired

SpiralFlex

Well-Known Member

Carrotsticks

Retired

SpiralFlex

Well-Known Member

Aysce

Well-Known Member

Carrotsticks

Retired

Aysce

Well-Known Member

Carrotsticks

Retired

Users Who Are Viewing This Thread (Users: 0, Guests: 1)