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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
avatar+109468 
+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.   frown

 

 Jan 29, 2019
edited by Melody  Jan 29, 2019
 #2
avatar+7824 
+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
avatar+109468 
+1

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

Melody  Jan 30, 2019

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