locus - ellipse

**afeasfaerw23231233** · June 4th 2008, 01:31 AM

p143 q9 c

I have problem in (c). I get the answer 2x+y = 0 but the book says it is 2x+3y = 0
question:
consider an ellipse E: $\frac {x^2}9 + \frac {y^2}4 = 1$
and a line L : $2x-3x+6=0$
and D: $y=mx+c$ is a line parallel to L.
a) if D cuts E at $Q(x_1 , y_1)$ and $R(x_2 , y_2)$ , show that $x_1$ and $x_2$ are the roots of the equation
$8x^2 +12cx + 9c^2- 36=0$
b) find the range of values of c if E and D cut at (i) 2 points / (ii) 1 point
c) M is the mid - point of QR. Find the equation of the locus M as c varies.

here's my working in (c)
$D: y = - \frac 2 3 x + c$
$x_1 + x_2 = - \frac {12c}8 = - \frac {3c}2$
$x = - \frac {3c}4$
$y_1 +y_2 = - \frac 2 3 (x_1 +x_2) + 2x = 3c$
$y= \frac 3 2 c = -2 x$
2x+y = 0
but answer says it's 2x+3y=0. what's wrong??? thanks!

**earboth** · June 4th 2008, 01:41 AM

Originally Posted by afeasfaerw23231233

p143 q9 c

I have problem in (c). I get the answer 2x+y = 0 but the book says it is 2x+3y = 0
question:
consider an ellipse E: $\frac {x^2}9 + \frac {y^2}4 = 1$
and a line L : $2x-3x+6=0$ I assume that you mean: 2x-3y+6=0
and D: $y=mx+c$ is a line parallel to L.
a) if D cuts E at $Q(x_1 , y_1)$ and $R(x_2 , y_2)$ , show that $x_1$ and $x_2$ are the roots of the equation
$8x^2 +12cx + 9c^2- 36=0$
b) find the range of values of c if E and D cut at (i) 2 points / (ii) 1 point
c) M is the mid - point of QR. Find the equation of the locus M as c varies.

here's my working in (c)
$D: y = - \frac 2 3 x + c$
$x_1 + x_2 = - \frac {12c}8 = - \frac {3c}2$
$x = - \frac {3c}4$
$y_1 +y_2 = - \frac 2 3 (x_1 +x_2) + 2x = 3c$
$y= \frac 3 2 c = -2 x$
2x+y = 0
but answer says it's 2x+3y=0. what's wrong??? thanks!

If you have $2x-3y+6=0~\iff~ 2x+6=3y~\iff~y=\frac23 x +2$

So you have used the wrong slope with D

**afeasfaerw23231233** · June 5th 2008, 05:36 AM

oh, no! i am so careless. thanks

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