The locus of the mid-points of the points of | vedclass

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

Question

(B) The given line is $x \cos 	heta + y \sin 	heta = 1$. To find the intersection points with the coordinate axes,we set $y = 0$ and $x = 0$ respect

Vedclass · Accepted Answer

$\frac{1}{x^2} + \frac{1}{y^2} = 4$

Vedclass · Answer

(B) The given line is $x \cos 	heta + y \sin 	heta = 1$. To find the intersection points with the coordinate axes,we set $y = 0$ and $x = 0$ respectively. For $y = 0$,$x = \frac{1}{\cos 	heta}$,so point $A = (\frac{1}{\cos 	heta}, 0)$. For $x = 0$,$y = \frac{1}{\sin 	heta}$,so point $B = (0, \frac{1}{\sin 	heta})$. Let $(h, k)$ be the mid-point of $AB$. Then $h = \frac{1}{2 \cos 	heta}$ and $k = \frac{1}{2 \sin 	heta}$. This implies $\cos 	heta = \frac{1}{2h}$ and $\sin 	heta = \frac{1}{2k}$. Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get $(\frac{1}{2h})^2 + (\frac{1}{2k})^2 = 1$. $\frac{1}{4h^2} + \frac{1}{4k^2} = 1$,which simplifies to $\frac{1}{h^2} + \frac{1}{k^2} = 4$. Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = 4$.

The locus of the mid-points of the points of intersection of $x \cos \theta + y \sin \theta = 1$ with the coordinate axes is

Similar Questions

$A(5,3), B(3,-2), C(2,-1)$ are three points. If $P(x,y)$ is a variable point such that the area of the quadrilateral $PABC$ is $10$ sq. units,then the locus of $P$ is

Let $A(2,1)$ be a point and the equation of the straight line $L$ be $x-y=0$. Let $a$ and $b$ respectively represent the distances from a variable point $P(\alpha, \beta)$ to $A$ and to the line $L$. If $c$ is the distance of the point $A$ from the origin such that $a=bc$,then the locus of $P$ is

$A$ variable line $L$ passing through the origin cuts two parallel lines $x-y+10=0$ and $x-y+20=0$ at two points $A$ and $B$ respectively. If $P$ is a point on line $L$ such that $OA, OP, OB$ are in harmonic progression,then the locus of $P$ is

Let $A = (a, 0)$ and $B = (-a, 0)$ be two fixed points. For $a \in (-\infty, 0)$,point $P(x, y)$ moves in the plane such that $PA = nPB$ $(n \neq 0, 1)$. If $n = 1$,find the locus of point $P$.

If $a, b, c$ are in harmonical progression,then the line $bcx + cay + ab = 0$ passes through a fixed point whose coordinates are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module