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

Answer:

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 .

Answer:

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) .

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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 .

Answer:

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