A rod of length 6 units slides with its ends | vedclass

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

Question

(A) Let $A(a, 0)$ and $B(0, b)$ be the end points of the rod and $P(h, k)$ be the midpoint.Since $P$ is the midpoint,we have $h = \frac{a+0}{2} = \fra

Vedclass · Accepted Answer

$x^2+y^2=9$

Vedclass · Answer

(A) Let $A(a, 0)$ and $B(0, b)$ be the end points of the rod and $P(h, k)$ be the midpoint.Since $P$ is the midpoint,we have $h = \frac{a+0}{2} = \frac{a}{2}$ and $k = \frac{0+b}{2} = \frac{b}{2}$.Thus,$a = 2h$ and $b = 2k$.The length of the rod is given as $6$ units,so $\sqrt{(a-0)^2 + (0-b)^2} = 6$.Squaring both sides,we get $a^2 + b^2 = 36$.Substituting the values of $a$ and $b$,we get $(2h)^2 + (2k)^2 = 36$.$4h^2 + 4k^2 = 36$.Dividing by $4$,we get $h^2 + k^2 = 9$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 9$.

$A$ rod of length $6$ units slides with its ends on the coordinate axes. The locus of the midpoint of the rod is

Similar Questions

The locus of the moving point $P$,such that $2PA = 3PB$ where $A$ is $(0,0)$ and $B$ is $(4,-3)$,is

If $P(1, 0)$,$Q(-1, 0)$,and $R(2, 0)$ are three given points,then find the locus of $S(x, y)$ which satisfies the relation $SQ^2 + SR^2 = 2SP^2$.

If a tangent to the circle $x^2 + y^2 = 1$ intersects the coordinate axes at distinct points $P$ and $Q,$ then the locus of the mid-point of $PQ$ is

The locus of the centre of the circle,which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and touches the line $x=2$,is

The centres of a set of circles,each of radius $2$,lie on the circle $x^2 + y^2 = 36$. The locus of any point in the set is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module