If the lines joining the origin to the points | vedclass

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

Question

(D) Given line $2 x+3 y=k \Rightarrow \frac{2 x+3 y}{k}=1 \dots (i)$Also given,$3 x^2-x y+3 y^2+2 x-3 y-4=0$Now homogenizing the above equation using

Vedclass · Accepted Answer

$6 k^2-5 k-52=0$

Vedclass · Answer

(D) Given line $2 x+3 y=k \Rightarrow \frac{2 x+3 y}{k}=1 \dots (i)$Also given,$3 x^2-x y+3 y^2+2 x-3 y-4=0$Now homogenizing the above equation using $(i)$:$3 x^2-x y+3 y^2+(2 x-3 y)\left(\frac{2 x+3 y}{k}\right)-4\left(\frac{2 x+3 y}{k}\right)^2=0$Multiplying by $k^2$:$k^2(3 x^2-x y+3 y^2) + k(4 x^2+6 x y-6 x y-9 y^2) - 4(4 x^2+9 y^2+12 x y) = 0$$x^2(3 k^2+4 k-16) + x y(-k^2-48) + y^2(3 k^2-9 k-36) = 0$Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(3 k^2+4 k-16) + (3 k^2-9 k-36) = 0$$6 k^2-5 k-52 = 0$

If the lines joining the origin to the points of intersection of $2 x+3 y=k$ and $3 x^2-x y+3 y^2+2 x-3 y-4=0$ are at right angles,then

Similar Questions

The equation of the pair of straight lines joining the origin to the points of intersection of the curve $x^2 + y^2 = 4$ and the line $y - x = 2$ is

The acute angle formed between the lines joining the origin to the points of intersection of the curves $x^2 + y^2 - 2x - 1 = 0$ and $x + y = 1$ is

If the lines joining the origin to the points of intersection of $y=mx+1$ and $x^2+y^2=1$ are perpendicular,then .........

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

The line $x+y=k$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k (>1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module