The equation of the locus of a point whose di | vedclass

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

Question

(B) Let the point be $P(h, k)$.Given that the distance of $P$ from $(a, 0)$ is equal to its distance from the $y$-axis.The distance of $P(h, k)$ from

Vedclass · Accepted Answer

$y^2 - 2ax + a^2 = 0$

Vedclass · Answer

(B) Let the point be $P(h, k)$.Given that the distance of $P$ from $(a, 0)$ is equal to its distance from the $y$-axis.The distance of $P(h, k)$ from $(a, 0)$ is $\sqrt{(h-a)^2 + (k-0)^2}$.The distance of $P(h, k)$ from the $y$-axis is $|h|$.Equating the distances: $\sqrt{(h-a)^2 + k^2} = |h|$.Squaring both sides: $(h-a)^2 + k^2 = h^2$.$h^2 - 2ah + a^2 + k^2 = h^2$.$k^2 - 2ah + a^2 = 0$.Replacing $(h, k)$ with $(x, y)$,the equation of the locus is $y^2 - 2ax + a^2 = 0$.

The equation of the locus of a point whose distance from $(a, 0)$ is equal to its distance from the $y$-axis is

Similar Questions

If the locus of the midpoints of the chords of the circle $x^2+y^2=25$,which subtend a right angle at the origin,is given by $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$,then $|a|=$

The locus of the point of intersection of perpendicular tangents to the circle $x^{2}+y^{2}=16$ is

The locus of the point of intersection of the straight lines $tx - 2y - 3t = 0$ and $x - 2ty + 3 = 0$ $(t \in R)$ is

If a circle passing through $A(1,1)$ touches the $X$-axis,then the locus of the other end of the diameter through $A$ is

The locus of a point which divides the join of $A(-1, 1)$ and a variable point $B$ on the circle ${x^2} + {y^2} = 4$ in the ratio $3 : 2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module