The ends of a rod of length l move on two mut | vedclass

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

Question

(D) Let the mutually perpendicular lines be the coordinate axes. Let the ends of the rod be $(a, 0)$ and $(0, b)$. Since the length of the rod is $l$,

Vedclass · Accepted Answer

$9x^2+36y^2=4l^2$

Vedclass · Answer

(D) Let the mutually perpendicular lines be the coordinate axes. Let the ends of the rod be $(a, 0)$ and $(0, b)$. Since the length of the rod is $l$,we have $a^2+b^2=l^2$. Let the point $(h, k)$ divide the rod in the ratio $1:2$. Using the section formula,we have: $h = \frac{1 \cdot 0 + 2 \cdot a}{1+2} = \frac{2a}{3} \Rightarrow a = \frac{3h}{2}$ $k = \frac{1 \cdot b + 2 \cdot 0}{1+2} = \frac{b}{3} \Rightarrow b = 3k$ Substituting these into the equation $a^2+b^2=l^2$: $(\frac{3h}{2})^2 + (3k)^2 = l^2$ $\frac{9h^2}{4} + 9k^2 = l^2$ $9h^2 + 36k^2 = 4l^2$ Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2+36y^2=4l^2$. Thus,option $(d)$ is correct.

The ends of a rod of length $l$ move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio $1:2$ is

Similar Questions

The locus of the mid-points of the perpendiculars drawn from points on the line $x=2y$ to the line $x=y$ is:

If $2a + b + 3c = 0$,then the line $ax + by + c = 0$ passes through the fixed point that is

Let $A (2, 3)$ and $B (-4, 5)$ be two fixed points. $A$ point $P$ moves in such a way that the area of $\Delta PAB = 12 \, \text{sq. units}$. Find its locus.

Let $A=(4,0)$ and $B=(0,12)$ be two points in the plane. The locus of a point $C(x, y)$ such that the area of $\triangle ABC$ is $18$ sq units is

Every point $(x, y)$ on the curve $3x + 2y - 3xy = 0$ is the centroid of a triangle formed by the coordinate axes and a line $(L)$ intersecting both the coordinate axes. Then all such lines $(L)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module