Find the locus of the midpoints of the chords | vedclass

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

Question

(A) The equation of the chord of a parabola $S = 0$ with midpoint $(h, k)$ is given by $T = S_1$.For the parabola $x^2 + 4y = 0$,the equation of the c

Vedclass · Accepted Answer

$x^2 + 2y + 2 = 0$

Vedclass · Answer

(A) The equation of the chord of a parabola $S = 0$ with midpoint $(h, k)$ is given by $T = S_1$.For the parabola $x^2 + 4y = 0$,the equation of the chord with midpoint $(h, k)$ is $xh + 2(y + k) = h^2 + 4k$.Since the chord passes through the focus $(0, -1)$,we substitute these coordinates into the equation:$x(0) + 2(-1 + k) = h^2 + 4k$$-2 + 2k = h^2 + 4k$$h^2 + 2k + 2 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + 2y + 2 = 0$.

Find the locus of the midpoints of the chords of the parabola $x^2 + 4y = 0$ which pass through its focus.

Similar Questions

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:

The slope of the chord of the parabola $y^2 = 4ax$ which passes through the origin and is normal at one of its endpoints is:

The vertex of a parabola is $(2, 2)$ and the coordinates of the two extremities of its latus rectum are $(-2, 0)$ and $(6, 0)$. The equation of the parabola is

If the focal distance of a point on the parabola $y^2 = x$ is $17/4$,and it lies in the first quadrant,find the equation of the normal at this point.

Let the image of the parabola $x^{2} = 4y$ in the line $x - y = 1$ be $(y + a)^{2} = b(x - c)$, where $a, b, c \in \mathbb{N}$. Then $a + b + c$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module