The locus of the orthocentre of the triangle formed by the lines

$ (1+p) x-p y+p(1+p)=0, $

$ (1+q) x-q y+q(1+q)=0,$

and $y=0$, where $p \neq q$, is

  • [IIT 2009]
  • A

    a hyperbola

  • B

    a parabola

  • C

    an ellipse

  • D

    a straight line

Similar Questions

The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is

The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is :

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are

  • [IIT 1979]

The vertex of an equilateral triangle is $(2,-1)$ and the equation of its base in $x + 2y = 1$. The length of its sides is

Let $\mathrm{A}(-2,-1), \mathrm{B}(1,0), \mathrm{C}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to_____.

  • [JEE MAIN 2024]