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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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.........

Similar Questions

The equation $2y^2 - xy - x^2 + 6x - 8 = 0$ represents

The area of the triangle formed by the lines $3x^2 - 4xy + y^2 = 0$ and $2x - y = 6$ is:

If the lines represented by the equation $2x^2 - 3xy + y^2 = 0$ make angles $\alpha$ and $\beta$ with the $x$-axis,then $\cot^2 \alpha + \cot^2 \beta = $

If one of the lines of the pair $ax^{2}+2hxy+by^{2}=0$ bisects the angle between the positive direction of the axes,then $a, b,$ and $h$ satisfy the relation:

If the equation $x^2 + y^2 + 2gx + 2fy + 1 = 0$ represents a pair of lines,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module