Question

The equation of parabola whose latus rectum is the line segment joining the points (–3, 1), (1, 1) is

• A. $(x+1{)}^2=4y$
• B. $(x-1{)}^2=4y$
• C. $(x+1{)}^2=2y$
• D. $(x+1{)}^2=4y$

Answer 1

The correct option is A $(x+1{)}^2=4y$

L(-3,1)$L^{\prime }$(1,1)
Mid point of $\left(LL\right){}^{\prime }$ isthe locus S(–1,1).
$4a=L{L}^{\prime }=4\Rightarrow a=1$
Axis is $\perp lrtoL{L}^{\prime }\Rightarrow$ Axis is parallel to y-axis
$\therefore$ Vertex A(–1, 0) or A (–1, 2) so that AS = a = 1
Parabola equation is $(x+1{)}^2=4y$ or $(x+1{)}^2=-4(y-2)$