Question

Equation of parabola having the extremities of its latus rectum as $(-4,1)$ and $(2,1)$ is

• A. $(x+1{)}^2=4(y-2)$
• B. $x^2=8y$
• C. $x^2=4(y-1)$
• D. None of these

Answer 1

The correct option is D None of these

The equation of the parabola is given as:

$(x-h{)}^2=4p(y-k)$

where, $4p$ is the length of latus rectum, which is equal to length of line segment joining the two given points $(-4,1)$ and $(2,1)$:

$4p=\sqrt{\left[\right(-4-2{)}^2+0]}=6$

and $p=\frac{6}{4}=\frac{3}{2}$

and focus is the mid point of the latus rectum, so focus: $(-1,1)$

the parabola is vertical, since latus rectum lies on $y=1$ line, and vertex $(h,k)=(-1,1+\frac{3}{2})=(-1,\frac{5}{2})$ ($p$ units up from $y-axis$)

using all these parameters, the equation of parabola will be

$(x+1{)}^2=6(y-\frac{5}{2})$