MCQ
If the points $(a{u^2},\;2au)$ and $(a{v^2},\;2av)$ are the extremities of a focal chord of the parabola ${y^2} = 4ax$, then
- A$uv - 1 = 0$
- ✓$uv + 1 = 0$
- C$u + v = 0$
- D$u - v = 0$
==> $y - 2au = \frac{{2av - 2au}}{{a{v^2} - a{u^2}}}(x - a{u^2})$
==> $y - 2au = \frac{{2a(v - u)}}{{a(v - u)(v + u)}}(x - a{u^2})$
==> $y - 2au = \frac{2}{{(v + u)}}(x - a{u^2})$
If this is focal chord, so it would passes through focus $(a,0)$.
==>$0 - 2au = \frac{2}{{v + u}}(a - a{u^2})$ ==>$ - uv - {u^2} = 1 - {u^2}$, $\therefore $ $uv+1=0.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
