The ends of the latus rectum of the parabola | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the parabola is $x^2 = -8y$.Comparing this with the standard form $x^2 = -4ay$,we get $4a = 8$,which implies $a = 2$.The foc

Vedclass · Accepted Answer

$(-4, -2)$ and $(4, -2)$

Vedclass · Answer

(C) The given equation of the parabola is $x^2 = -8y$.Comparing this with the standard form $x^2 = -4ay$,we get $4a = 8$,which implies $a = 2$.The focus of the parabola $x^2 = -4ay$ is $(0, -a)$,so the focus is $(0, -2)$.The latus rectum is a line parallel to the $x$-axis passing through the focus $(0, -2)$,given by the equation $y = -2$.To find the ends of the latus rectum,substitute $y = -2$ into the parabola equation $x^2 = -8y$:$x^2 = -8(-2) = 16$$x = \pm 4$.Thus,the ends of the latus rectum are $(4, -2)$ and $(-4, -2)$.

The ends of the latus rectum of the parabola $x^2 + 8y = 0$ are

Similar Questions

The line among the following which touches the parabola $y^2=4ax$ is

The angle between the tangents drawn to the parabola $y^{2}=12x$ from the point $(-3, 2)$ is (in $^{\circ}$)

Find the equation of the parabola that satisfies the following conditions: Focus $(6, 0)$,directrix $x = -6$.

The smallest value of $x^2 - 3x + 3$ in the interval $(-3, 3/2)$ is

If the vertex of the conic $y^{2}-4y=4x-4a$ always lies between the straight lines $x+y=3$ and $2x+2y-1=0$, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module