Homogenization

Homogenize $y^2=x$ with $x+y=4$.

Homogenize $y^2=x$ with $x+y=3$.

Homogenize $y^2=x$ with $x+y=2$.

Find the angle between the lines joining the origin to the points of intersection of $y^2=x$ and $x+y=1$.

The equation of lines joining origin and points of intersection of curves ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}\mathrm{y}=1$ and $3{\mathrm{x}}^2+5{\mathrm{y}}^2–\mathrm{ }\mathrm{x}\mathrm{y}=11$ is

The equation of lines joining origin and points of intersection of curves ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}\mathrm{y}=1$ and $3{\mathrm{x}}^2+5{\mathrm{y}}^2–\mathrm{ }\mathrm{x}\mathrm{y}=11$ is

If the line $x+2y=k$, intersects the curve $x^2-xy+y^2+3x+3y-2=0$ at two points $A$ and $B$ and if $O$ is the origin, then the condition for $AOB=90°$ is

The feet of the perpendicular from the origin on a variable chord of the circle $x^2+y^2-2x-2y=0$ is $N$. If the variable chord makes an angle of ${90}^{°}$ at the origin, then the equation of circle having centre at $(-1,-2)$ and intersects the locus of $N$ orthogonally is:

If the line $\frac{x-1}{15}=\frac{y+1}{16}=\frac{z-1}{n}$ intersect the curve $6x^2+5y^2=1, z=0$ then $10n^2-20n+k=0$ where the value of $k$ is

The angle between the straight line joining the point $\left(–1, 0\right)$ to the common points of

$3x^2+5xy-3y^2+8x+8y+5=0$ and $3x-2y+2 =0$ is

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

The lines joining the origin to the points of intersection of the line $3x-2y=1$ and the curve $3x^2+5xy-3y^2+2x+3y=0,$ are

The lines joining the points of intersection of the curve ${\left(x-h\right)}^2+{\left(y-k\right)}^2-c^2=0$ and the line $kx+hy=2hk$ to the origin are perpendicular, then

The lines joining the points of intersection of curve $5x^2+12xy-8y^2+8x-4y+12=0$ and the line $x-y=2$ to the origin, makes the angles with the axes

The lines joining the points of intersection of line $x+y=1$ and curve $x^2+y^2-2y+\lambda =0$ to the origin are perpendicular, then the value of $\lambda$ will be

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

If the lines joining origin to the points of intersection of the line $fx-gy=\lambda$ and the curve $x^2+hxy-y^2+gx+fy=0$ be mutually perpendicular, then

The lines joining the origin to the points of intersection of the curves $a{x}^2+2hxy +b{y}^2+2gx=0$ and $d{x}^2+2h^{\prime }xy+b^{\prime }{y}^2+2g^{\prime }x=0$ will be mutually perpendicular, if

The lines joining the origin to the points of intersection of the line $3x-2y=1$ and the curve $3x^2+5xy-3y^2+2x+3y=0,$ are

The lines joining the points of intersection of the curve ${\left(x-h\right)}^2+{\left(y-k\right)}^2-c^2=0$ and the line $kx+hy=2hk$ to the origin are perpendicular, then