PQ is a double ordinate of the parabola y^2 = | vedclass

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

Question

(A) Let the coordinates of $P$ be $(h, k)$ and $Q$ be $(h, -k)$. Since $P$ lies on the parabola $y^2 = 4ax$,we have $k^2 = 4ah$.Let $(x, y)$ be a poin

Vedclass · Accepted Answer

$9y^2 = 4ax$

Vedclass · Answer

(A) Let the coordinates of $P$ be $(h, k)$ and $Q$ be $(h, -k)$. Since $P$ lies on the parabola $y^2 = 4ax$,we have $k^2 = 4ah$.Let $(x, y)$ be a point of trisection of $PQ$. The points of trisection divide $PQ$ in the ratio $1:2$ or $2:1$.For the point dividing $PQ$ in ratio $1:2$,the $y$-coordinate is $y = \frac{1(-k) + 2(k)}{1+2} = \frac{k}{3}$,so $k = 3y$.The $x$-coordinate is $x = h$.Substituting $k = 3y$ and $h = x$ into the parabola equation $k^2 = 4ah$,we get $(3y)^2 = 4ax$,which simplifies to $9y^2 = 4ax$.

$PQ$ is a double ordinate of the parabola $y^2 = 4ax$. The locus of the points of trisection of $PQ$ is

Similar Questions

Find the equation of the normal to the parabola $y^2 = 8x$ at the point $(2, 4)$.

The locus of a point which divides the line segment joining the point $(0,-1)$ and a point on the parabola $x^{2}=4y$ internally in the ratio $1:2$ is:

If $P$ is $(3, 1)$ and $Q$ is a point on the curve $y^2 = 8x$,then the locus of the mid-point of the line segment $PQ$ is

The tangents at the points $A(1, 3)$ and $B(1, -1)$ on the parabola $y^{2} - 2x - 2y = 1$ meet at the point $P$. Then the area (in unit$^{2}$) of the triangle $PAB$ is:

If the equation of the parabola,whose vertex is at $(5,4)$ and the directrix is $3x+y-29=0$,is $x^{2}+ay^{2}+bxy+cx+dy+k=0$,then $a+b+c+d+k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module