If the equation 8x^2+8xy+2y^2+26x+13y+15=0 re | vedclass

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(B) The given equation is $8x^2+8xy+2y^2+26x+13y+15=0$ ...$(i)$We can rewrite the equation as $2(4x^2+4xy+y^2)+13(2x+y)+15=0$.Let $2x+y=t$. Then the e

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$\frac{7}{2\sqrt{5}}$

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(B) The given equation is $8x^2+8xy+2y^2+26x+13y+15=0$ ...$(i)$We can rewrite the equation as $2(4x^2+4xy+y^2)+13(2x+y)+15=0$.Let $2x+y=t$. Then the equation becomes $2t^2+13t+15=0$.Factoring the quadratic: $2t^2+10t+3t+15=0 \Rightarrow 2t(t+5)+3(t+5)=0$.So,$(2t+3)(t+5)=0$.Substituting $t=2x+y$ back,we get $(2(2x+y)+3)(2x+y+5)=0$.This simplifies to $(4x+2y+3)(2x+y+5)=0$.Thus,the two lines are $4x+2y+3=0$ and $2x+y+5=0$.Dividing the first line by $2$,we get $2x+y+1.5=0$.The distance $d$ between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.Here,$A=2, B=1, C_1=5, C_2=1.5$.$d = \frac{|5-1.5|}{\sqrt{2^2+1^2}} = \frac{3.5}{\sqrt{5}} = \frac{7}{2\sqrt{5}}$.

If the equation $8x^2+8xy+2y^2+26x+13y+15=0$ represents a pair of parallel straight lines,then the distance between them is.........

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