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# help

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Let a and b be real numbers such that the equation 6x^2 - axy - 3y^2 - 24x + 3y + b = 0 represents two lines, whose intersection lies on the x -axis. Enter the ordered pair (a,b).

Jan 29, 2019

#1
+102763
+3

Let a and b be real numbers such that the equation 6x^2 - axy - 3y^2 - 24x + 3y + b = 0 represents two lines, whose intersection lies on the x -axis. Enter the ordered pair (a,b).

If the intersection is on the x axis then that is when y=0

sub in y=0 and I get

$$6x^2 - axy - 3y^2 - 24x + 3y + b = 0\\ when\;\;y=0\\ 6x^2 - 24x + b = 0\\ x=\frac{24\pm\sqrt{24^2-24b}}{12}\\ \text{if there is only one solution then}\\ 24^2-24b=0\\ 24-b=0\\ b=24$$

if b=24 then we have

$$6x^2 - axy - 3y^2 - 24x + 3y + 24 = 0$$

I have used a Desmos graph to determine that a=1.5 approx

So    (a,b) = (1.5,24)

BUT I have not worked out how to get the value of a without using the graphing calculator.

My knowledge of conics is not great.

Jan 29, 2019
edited by Melody  Jan 29, 2019
#2
+7709
+4

Define 2 linear functions: $$L_1:Ax+By+C=0\\ L_2:Dx+Ey+F=0$$.

From the question: The 2 straight lines intersect at the x-axis. $$\Rightarrow \dfrac{C}{A} = \dfrac{F}{D}$$

As C/A = F/D, C = kF and A = kD for some real k. We can rewrite L2 as: $$L_2:Ax+ny+C=0$$.

The equation representing the 2 straight lines is

$$A^2x^2+(An+AB)xy+Bny^2+2ACx+(BC+Cn)y+C^2=0$$.

Define $$n + B \stackrel{\text{def}}{=} P$$.

Now the rewritten equation is $$A^2x^2+APxy+Bny^2+2ACx+CPy+C^2=0$$.

WLOG, we can assume A > 0.

In this case, A = $$\sqrt6$$.

2AC = -24, so C = $$-2\sqrt6$$. => b = C2 = 24

CP = 3, so P =  $$-\dfrac{\sqrt6}{4}$$.

-a = AP = -3/2

So a = 3/2 :D

There we get (a,b) = (3/2, 24). :D

Jan 29, 2019
edited by MaxWong  Jan 29, 2019
#3
+102763
+1

Thanks Max, I am really pleased that you have answered this :))

Melody  Jan 30, 2019