If 9x^2-24xy+16y^2+alpha x+beta y+6=0 represe | vedclass

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

Question

(D) The given equation is $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$. This can be written as $(3x-4y)^2+\alpha x+\beta y+6=0$. Since it represents paralle

Vedclass · Accepted Answer

$\frac{-3}{4}$

Vedclass · Answer

(D) The given equation is $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$. This can be written as $(3x-4y)^2+\alpha x+\beta y+6=0$. Since it represents parallel lines,let the lines be $(3x-4y+k_1)=0$ and $(3x-4y+k_2)=0$. Their product is $(3x-4y+k_1)(3x-4y+k_2) = 9x^2-24xy+16y^2+3(k_1+k_2)x-4(k_1+k_2)y+k_1k_2=0$. Comparing coefficients: $\alpha = 3(k_1+k_2)$,$\beta = -4(k_1+k_2)$,and $k_1k_2 = 6$. Thus,$\frac{\alpha}{\beta} = \frac{3(k_1+k_2)}{-4(k_1+k_2)} = -\frac{3}{4}$. The distance between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$. Here,$\frac{|k_1-k_2|}{\sqrt{3^2+(-4)^2}} = 1 \implies |k_1-k_2| = 5$. Since $(k_1-k_2)^2 = (k_1+k_2)^2 - 4k_1k_2$,we have $25 = (k_1+k_2)^2 - 24$,so $(k_1+k_2)^2 = 49$,meaning $k_1+k_2 = \pm 7$. If one line passes through $(1,1)$,then $3(1)-4(1)+k = 0 \implies k = 1$. If $k_1=1$,then $k_2 = 6$ (since $k_1k_2=6$),so $k_1+k_2 = 7$. Then $\alpha = 3(7) = 21$ and $\beta = -4(7) = -28$. Thus,$\frac{\alpha}{\beta} = \frac{21}{-28} = -\frac{3}{4}$.

If $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$ represents a pair of parallel lines $1$ unit apart and one of those lines passes through $(1,1)$,then $\frac{\alpha}{\beta} = $

Similar Questions

If the equations of the opposite sides of a parallelogram are $x^2 - 7x + 6 = 0$ and $y^2 - 14y + 40 = 0$,then the equation of one of its diagonals is

If the pairs of straight lines represented by $3x^2+2hxy-3y^2=0$ and $3x^2+2hxy-3y^2+2x-4y+c=0$ form a square,then $(h, c) =$

The area (in sq units) of the quadrilateral formed by two pairs of lines $\lambda^2 x^2 - m^2 y^2 - n(\lambda x + my) = 0$ and $\lambda^2 x^2 - m^2 y^2 + n(\lambda x + my) = 0$ is:

The area of the triangle formed by the pair of straight lines $(ax+by)^2 - 3(bx-ay)^2 = 0$ and the line $ax+by+c = 0$ is

If $2x^2+3xy-2y^2=0$ represents two sides of a parallelogram and $3x+y+1=0$ is one of its diagonals,then the other diagonal is

Vedclass Test Series

Exam Paper Generator

Online Exam Module