(A) Given $A = (a, 0)$ and $B = (-a, 0)$.Let $P = (x, y)$.The condition is $PA = nPB$.For $n = 1$,we have $PA = PB$.Squaring both sides,$PA^2 = PB^2$.

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(A) Given $A = (a, 0)$ and $B = (-a, 0)$.Let $P = (x, y)$.The condition is $PA = nPB$.For $n = 1$,we have $PA = PB$.Squaring both sides,$PA^2 = PB^2$.$(x - a)^2 + (y - 0)^2 = (x - (-a))^2 + (y - 0)^2$.$(x - a)^2 + y^2 = (x + a)^2 + y^2$.$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2 + y^2$.$-2ax = 2ax$.$4ax = 0$.Since $a \neq 0$ (as $a \in (-\infty, 0)$),we get $x = 0$.The equation $x = 0$ represents the $y$-axis,which is a straight line.

Class 11 Mathematics Straight Line Locus of Point

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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 $P$.

A
$A$ straight line
B
$A$ circle
C
$A$ parabola
D
$A$ hyperbola

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Straight Line

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Locus of Point

The portion of the tangent to the curve $x^{2/3} + y^{2/3} = a^{2/3}, a > 0$ at any point, intercepted between the axes, is:

MediumWBJEE 2023

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DifficultJEE MAIN 2019

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Consider a square $ABCD$ of side $12$ and let $M, N$ be the midpoints of $AB, CD$ respectively. Take a point $P$ on $MN$ and let $AP=r, PC=s$. Then,the area of the triangle whose sides are $r, s, 12$ is

DifficultKVPY 2010

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$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is

MediumAP EAMCET 2017

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If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point:

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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 $P$.

Similar Questions

The portion of the tangent to the curve $x^{2/3} + y^{2/3} = a^{2/3}, a > 0$ at any point, intercepted between the axes, is:

$A$ point $P$ moves on the line $2x - 3y + 4 = 0$. If $Q(1, 4)$ and $R(3, -2)$ are fixed points,then the locus of the centroid of $\Delta PQR$ is a line

Consider a square $ABCD$ of side $12$ and let $M, N$ be the midpoints of $AB, CD$ respectively. Take a point $P$ on $MN$ and let $AP=r, PC=s$. Then,the area of the triangle whose sides are $r, s, 12$ is

$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is

If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point:

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