Question 1: Find the locus of a point equidistant from the point $(2,4)$ and the y-axis.

Let $P(h, k)$ be the point which is equidistant from point $( 2, 4)$ and y-axis.

Distant of point $( h, k)$ from y-axis is $h$.

$\therefore h = \sqrt{ (h-2)^2 + ( k-4)^2}$

$\Rightarrow h^2 = (h-2)^2 + ( k-4)^2$

$\Rightarrow h^2 = h^2 + 4 - 4 h + k^2 + 16 - 8k$

$\Rightarrow k^2 - 4h - 8k + 20 = 0$

Hence the locus of $(h, k)$ is $y^2 - 4x-8y + 20 = 0$

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Question 2: Find the equation of the locus of a point which moves such that the ratio of its distances from $(2,0)$ and $(1,3)$ is $5 :4$.

Given $A( 2, 0), B( 1, 3)$. Let $P(h,k)$ be the point which moves such that the ratio of its distances from $(2,0)$ and $(1,3)$ is $5 :4$

Given $\frac{PA}{PB}$ $=$ $\frac{5}{4}$

$\frac{\sqrt{ (h-2)^2 + ( k-0)^2}}{\sqrt{ (h-1)^2 + ( k-3)^2}}$ $=$ $\frac{5}{4}$

$\Rightarrow 16 ( h^2 + 4 - 4h + k^2) = 25 ( h^2 + 1 - 2h +k^2 + 9 - 6k)$

$\Rightarrow 16h^2 +64 - 64 h + 16k^2 = 25h^2 - 50h + 25k^2 + 250 - 150k$

$\Rightarrow 9h^2 + 9k^2 + 14h - 150 k + 186 = 0$

Hence the locus of $(h, k)$ is $9x2 + 9y^2 + 14x - 150 y + 186 = 0$

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Question 3: A point moves as so that the difference of its distances from $(ae,0)$ and $(-ae,0)$ is $2a$, prove that the equation to its locus $b$ $\frac{x^2}{a}$ $-$ $\frac{y^2}{b}$ $= 1$, where $b^2 = a^2 ( e^2 -1)$.

Given $A(ae,0)$ and $B(-ae,0)$ is $2a$. Let $P(h, k)$ be the point.

$\therefore PA- PB = 2a$

$\sqrt{(h-ae)^2+(k-0)^2} - \sqrt{(h+ae)^2+(k-0)^2} = 2a$

$\sqrt{(h-ae)^2+k^2} = 2a + \sqrt{(h+ae)^2+k^2}$

Squaring both sides we get

$(h-ae)^2+k^2 = 4a^2 + (h+ae)^2 + k^2 + 4a \sqrt{(h+ae)^2+k^2}$

$h^2 + a^2e^2- 2aeh + k^2 = 4a^2 + h^2 + a^2e^2 + 2aeh + k^2 + 4a \sqrt{(h+ae)^2+k^2}$

$-4aeh-4a^2 = 4a \sqrt{(h+ae)^2+k^2}$

$-eh-a = \sqrt{(h+ae)^2+k^2}$

Squaring once again we get

$e^2 h^2 + a^2 + 2aeh = h^2 + a^2 e^2 + 2aeh + k^2$

$h^ ( e^2 - 1) = a^2 ( e^2 - 1) + k^2$

$h^2 = a^2 +$ $\frac{k^2}{e^2 - 1}$

$\frac{h^2}{a^2}$ $-$ $\frac{k^2}{a^2(e^2-1)}$ $= 1$

$\frac{h^2}{a^2}$ $-$ $\frac{k^2}{b^2}$ $= 1$

Hence locus of $(h, k)$ is $\frac{h^2}{x^2}$ $-$ $\frac{y^2}{b^2}$ $= 1$ where $b^2 = a^2(e^2-1)$

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Question 4: Find the locus of a point such that the sum of its distances from $(0,2)$ and $(0, -2)$ is $6$.

Given $A(0,2)$ and $B(0, -2)$. Let $P(h,k)$ be the point such that the sum of its distances from $(0,2)$ and $(0, -2)$ is $6$.

$\therefore \sqrt{ (h-0)^2+(k-2)^2} + \sqrt{ (h-0)^2+(k+2)^2} = 6$

$\Rightarrow \sqrt{ h^2+(k-2)^2} = 6 - \sqrt{ h^2+(k+2)^2}$

Squaring both sides

$h^2 + k^2 + 4 - 4k = 36 + h^2 + k^2 + 4 + 4k - 12\sqrt{ h^2+(k+2)^2}$

$\Rightarrow -36 -8k = 12 \sqrt{ h^2+(k+2)^2}$

$\Rightarrow -9 -2k = 3 \sqrt{ h^2+(k+2)^2}$

Squaring once again we get

$\Rightarrow 81+4k^2 + 36k = 9 ( h^2 + k^2 + 4 + 4k)$

$\Rightarrow 81+4k^2 + 36k = 9h^2 + 9k^2 + 36 + 36k$

$\Rightarrow 9h^2 + 5k^2 - 45 = 0$

Hence the locus of $(h,k)$ is $9x^2 + 5y^2 - 45 = 0$

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Question 5: Find the locus of a point which is equidistant from $(1, 3)$ and x-axis.

Given $A(1,3)$ and x-axis. Let $P(h,k)$ be the point.

$AP = k \Rightarrow AP^2 = k^2$

$\Rightarrow (h-1)^2 + (k-3)^2 = k^2$

$\Rightarrow h^2 + 1 - 2h + k^2 + 9 -6k = k^2$

$\Rightarrow h^2 -2h - 6k +10 = 0$

Hence the locus of $(h,k)$ is $x^2 -2x - 6y +10 = 0$

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Question 6: Find the locus of a point which moves such that its distance from the origin is three times its distance from x-axis.

Let $P(h,k)$ be the point. The distance of $P(h, k)$ from x-axis is $k$

$\therefore OP = k \Rightarrow OP^2 = k^2$

$\Rightarrow (h-0)^2 + ( k-0)^2 = ( 3k)^2$

$\Rightarrow h^2 + k^2 = 9k^2$

$\Rightarrow h^2 = 8k^2$

Hence the locus of $(h, k)$ is $x^2 = 8y^2$

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Question 7: $A (5, 3), B (3, - 2)$ are two fixed points; find the equation to the locus of a point $P$ which moves so that the area of the $\triangle PAB$ is $9$ units.

Given $A (5, 3), B (3, - 2)$. Let $P(h,k)$ be the point.

Area of $\triangle ABP = \frac{1}{2} \{ x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3( y_1 - y_2) \}$

$\Rightarrow 9 = \frac{1}{2} | 5( -2-k) + 3(k-3) + h ( 3+2) |$

$\Rightarrow | 5h - 2k - 19 | = 18$

Therefore $5h - 2k - 19 = 18$      or      $5h - 2k - 19 = -18$

$\Rightarrow 5h -2k - 37 = 0$      or      $5h-2k-1=0$

Hence the locus of $(h, k) is 5x-2y - 37 = 0$      or      $5x - 2y - 1 = 0$

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Question 8: Find the locus of a point such that the line segments having end points $(2, 0)$ and $(-2, 0)$ subtend a right angle at that point.

Given $(2, 0)$ and $(-2, 0)$. Let $P(h,k)$ be the point such that $\angle APB = 90^{\circ}$

$\therefore AB^2 = AP^2 + BP^2$

$\Rightarrow ( 2+2)^2 = ( h - 2)^2 + k^2 + ( h+2)^2 + k^2$

$\Rightarrow 16 = h^2 + 4 - 4h + k^2 + h^2 + 4 + 4h + k^2$

$\Rightarrow 8 = 2h^2 + 2k^2$

$\Rightarrow h^2 + k^2 = 4$

Hence the locus of $(h, k)$ is $x^2 + y^2 = 4$

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Question 9: If $A (-1, 1)$ and $B (2, 3)$ are two fixed points, find the locus of a point $P$ so that the area of $\triangle PAB=8$ sq. units.

Given $A (-1, 1)$ and $B (2, 3)$. Let $P(h,k)$ be the point.

Therefore are of $\triangle PAB = \frac{1}{2} \{ x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3( y_1 - y_2) \}$

$\Rightarrow 8 \times 2 = | -1( 3-k) + 2 ( k-1) + h ( 1 - 3) |$

$\Rightarrow 16 = | 2h - 3k + 5|$

$\therefore 2h - 3k + 5 = 16$      or      $2h - 3k + 5 = -16$

$\Rightarrow 2h - 3k -11 = 0$      or      $2h - 3k + 17 = 0$

Hence the locus of $( h, k)$ is $2x - 3y - 11 = 0$      or    $2x-3y+17 = 0$

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Question 10: A rod of length $l$ slides between the two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio $7 :2$.

Let $x$ and $y$ axis be the two perpendicular lines.

Let $A(a,0)$ and $B ( 0, b)$ be the two intercepts on the axes.

Let $P(h, k)$ be the point that divides $AB$ in $1:2$ ratio.

$\therefore$ $\frac{BP}{AP}$ $=$ $\frac{1}{2}$

$\therefore ( h, k) = \Big($ $\frac{a+0}{3}$ $,$ $\frac{0+2b}{3}$ $\Big ) = \Big($ $\frac{a}{3}$ $,$ $\frac{2b}{3}$ $\Big)$

$\Rightarrow a = 3h$ and $h =$ $\frac{3k}{2}$

Now $AB = l$

$\therefore \sqrt{a^2 + b^2} = l$

$\Rightarrow a^2 + b^2 = l ^2$

$\Rightarrow (3h)^2 + \Big($ $\frac{3k}{2}$ $\Big )^2 = l^2$

$h^2 +$ $\frac{k^2}{4}$ $=$ $\frac{l^2}{9}$

Hence locus of $(h, k)$ is $x^2 +$ $\frac{y^2}{4}$ $=$ $\frac{l^2}{9}$

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Question 11: Find the locus bf the mid-point of the portion of the line $x \cos \alpha + y sin \alpha = p$ which is intercepted between the axes.

Given line is $x \cos \alpha + y \sin \alpha = p$

When $x = 0 , y =$ $\frac{p}{\sin \alpha}$ $= p \mathrm{cosec} \alpha$

When $y = 0, x =$ $\frac{p}{\cos \alpha}$ $= p \sec \alpha$

Let $P(h, k)$ be th emid point of $AB$

$\therefore h =$ $\frac{p \sec \alpha+0}{2}$           $h =$ $\frac{0 + p \mathrm{cosec} \alpha}{2}$

We know $\cos^2 \alpha + \sin^2 \alpha = 1$

$\Rightarrow$ $\frac{p^2}{4h^2}$ $+$ $\frac{p^2}{4k^2}$ $= 1$

$\Rightarrow$ $\frac{1}{h^2}$ $+$ $\frac{1}{k^2}$ $=\frac{4}{p^2}$

Hence the locus of $(h, k)$ is $\frac{1}{x^2}$ $+$ $\frac{1}{y^2}$ $=$ $\frac{4}{p^2}$

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Question 12: If $O$ is the origin and $Q$ is a variable point on $y^2 =x$. Find the locus of the mid-point of $OQ$.

Let $Q$ be $( a, b)$. It lies on $y^2 = x$

$\Rightarrow b^2 = a$     … … … … … i)

Let $P(h,k)$ be the mid point of $OQ$

$\therefore h =$ $\frac{0+a}{2}$     and     $k =$ $\frac{0+b}{2}$

$\Rightarrow a = 2h$    and      $b = 2k$

Substituting in i) we get $( 2k)^2 = 2h \Rightarrow 2k^2 = h$

Hence the locus of $( h, k)$ is $2y^2 = x$