If the pair of lines joining the origin and t | vedclass

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

Question

(C) The equation of the line is $ax+by=1$,which can be written as $ax+by=1$. The equation of the curve is $x^2+y^2-x-y-1=0$.By homogenizing the equati

Vedclass · Accepted Answer

$\sqrt{5/2}$

Vedclass · Answer

(C) The equation of the line is $ax+by=1$,which can be written as $ax+by=1$. The equation of the curve is $x^2+y^2-x-y-1=0$.By homogenizing the equation of the curve using the line equation,we get:$x^2+y^2-(x+y)(ax+by)-(ax+by)^2=0$$x^2+y^2-(ax^2+bxy+axy+by^2)-(a^2x^2+b^2y^2+2abxy)=0$$x^2(1-a-a^2)+xy(-a-b-2ab)+y^2(1-b-b^2)=0$Since the pair of lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(1-a-a^2)+(1-b-b^2)=0$$a^2+b^2+a+b-2=0$This represents a circle in the $(a, b)$ plane. Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we have $g=1/2$,$f=1/2$,and $c=-2$.The radius $r$ is given by $\sqrt{g^2+f^2-c}$:$r = \sqrt{(1/2)^2+(1/2)^2-(-2)}$$r = \sqrt{1/4+1/4+2} = \sqrt{1/2+2} = \sqrt{5/2}$.

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

Similar Questions

The lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ will be equidistant from the origin,if

Let $L$ be the line joining the origin to the point of intersection of the lines represented by $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. If $L$ is perpendicular to the line $kx + y + 3 = 0$,then $k$ is equal to

If the angle between the lines joining the origin to the points of intersection of $x+2y+\lambda=0$ and $2x^2-2xy+3y^2+2x-y-1=0$ is $\frac{\pi}{2}$,then a value of $\lambda$ is

The distance between the two parallel lines represented by the equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$ is

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module