$12x^2 +7xy -py^2-18x+qy+6=0$ represents a pair of $\perp$ straight lines. Find $p$ and $q$.

Question

I'm sorry I can't show you what I've tried because I don't know how to start. For those who start from behind: p=12,q=1.

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Any help would be appreciated. :) thanks.

Answer 1

As the given equation needs to represent a pair of perpendicular straight lines,

let $ax+by+c=0, bx-ay+d=0$

$$(ax+by+c)(bx-ay+d)=ab x^2+(b^2-a^2)xy-aby^2+x(ad+bc)+y(bd-ca)+cd$$

Comparing with the given equation

$ab=12,p=ab=12$(done)

$cd=6,,b^2-a^2=7,q=bd-ca,-18=ad+bc$

$$b^2+a^2=\sqrt{(a^2-b^2)^2+(2ab)^2}=25\implies b^2=16,a^2=9$$

As $ab=12>0,a,b$ must have same sign

Taking $b=4,a=3;3d+4c=-18,cd =6$ Can you solve for $c,d?$

Then $q=bd-ca$