The line x+2y=k meets the curve 2x^2-2xy+3y^2 | vedclass

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

Question

(A) Given the line $x+2y=k$,we have $\frac{x+2y}{k}=1$. Substituting this into the equation of the curve $2x^2-2xy+3y^2+(2x-y)(1)-(1)^2=0$,we get: $2x

Vedclass · Accepted Answer

$\pm 1$

Vedclass · Answer

(A) Given the line $x+2y=k$,we have $\frac{x+2y}{k}=1$. Substituting this into the equation of the curve $2x^2-2xy+3y^2+(2x-y)(1)-(1)^2=0$,we get: $2x^2-2xy+3y^2+(2x-y)\left(\frac{x+2y}{k}\right)-\left(\frac{x+2y}{k}\right)^2=0$. Multiplying by $k^2$,we obtain: $k^2(2x^2-2xy+3y^2)+k(2x^2+4xy-xy-2y^2)-(x^2+4xy+4y^2)=0$. Expanding and grouping the terms: $x^2(2k^2+2k-1) - xy(2k^2-3k+4) + y^2(3k^2-2k-4) = 0$. Since $OA \perp OB$,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(2k^2+2k-1) + (3k^2-2k-4) = 0$. $5k^2 - 5 = 0$. $k^2 = 1$. Thus,$k = \pm 1$.

The line $x+2y=k$ meets the curve $2x^2-2xy+3y^2+2x-y-1=0$ at two points $A$ and $B$. Let $O$ be the origin. If the line segments $OA$ and $OB$ are perpendicular to each other,then $k=$

Similar Questions

The acute angle between the pair of straight lines joining the origin to the points of intersection of the line $x+y-1=0$ with the pair of straight lines $k x^2+8 x y-3 y^2+2 x-4 y-1=0$ is

The condition that the lines joining the origin to the points of intersection of the line $\frac{x}{a} + \frac{y}{b} = 2$ and the circle $(x - a)^2 + (y - b)^2 = r^2$ are at right angles is

The distance between the parallel lines given by $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$ is

The distance between the two lines represented by the pair of straight lines $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$ is:

The lines joining the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$ to the origin are perpendicular. Then the value of $\lambda$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module