The equation to the locus of a point which mo | vedclass

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

Question

(B) Let the moving point be $P(x, y)$.The distance of point $P(x, y)$ from the $x$-axis is $|y|$.The distance of point $P(x, y)$ from the origin $(0,

Vedclass · Accepted Answer

$x^2 - 3y^2 = 0$

Vedclass · Answer

(B) Let the moving point be $P(x, y)$.The distance of point $P(x, y)$ from the $x$-axis is $|y|$.The distance of point $P(x, y)$ from the origin $(0, 0)$ is $\sqrt{x^2 + y^2}$.According to the given condition,the distance from the $x$-axis is one-half of its distance from the origin:$|y| = \frac{1}{2} \sqrt{x^2 + y^2}$Squaring both sides,we get:$y^2 = \frac{1}{4} (x^2 + y^2)$$4y^2 = x^2 + y^2$$x^2 + y^2 - 4y^2 = 0$$x^2 - 3y^2 = 0$Thus,the required equation of the locus is $x^2 - 3y^2 = 0$.

The equation to the locus of a point which moves so that its distance from the $x$-axis is always one-half its distance from the origin is:

Similar Questions

$A$ cow is tied to a corner (vertex) of a regular hexagonal fenced area of side $a \ m$ by a rope of length $\frac{5a}{2} \ m$ in a grass field. (The cow cannot graze inside the fenced area). What is the maximum possible area of the grass field to which the cow has access to graze?

The locus of the centre of the circle which cuts off intercepts of length $2a$ and $2b$ from the $x$-axis and $y$-axis respectively,is

$A$ point $P(x, y)$ is such that the sum of the squares of its distances from $(a, 0)$ and $(-a, 0)$ is $2b^2$. The equation representing the locus of $P$ is

The locus of the centre of a circle touching both the coordinate axes is

If a circle passes through the point $(a, b)$ and cuts the circle $x^2 + y^2 = K^2$ orthogonally,then the equation of the locus of its centre is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module