Class 11: Cartesian System of Rectangular Co-ordinates – Exercise 22.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

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

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

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

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

$\displaystyle \text{Hence locus of } (h, k) \text{ or } y^2 - 4x-8y + 20 = 0$

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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 $\displaystyle (2,0) \text{ and } (1,3) \text{ or } 5 :4$ .

Answer:

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

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

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

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

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

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

$\displaystyle \text{Hence locus of } (h, k) \text{ or } 9x2 + 9y^2 + 14x - 150 y + 186 = 0$

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

Answer:

$\displaystyle \text{Given } A(ae,0) \text{ and } B(-ae,0) \text{ or } 2a . \text{ Let } P(h, k)$ be the point.

$\displaystyle \therefore PA- PB = 2a$

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

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

Squaring both sides we get

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

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

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

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

Squaring once again we get

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

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

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

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

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

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

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Question 4: Find the locus of a point such that the sum of its distances from $\displaystyle (0,2) \text{ and } (0, -2) \text{ or } 6$ .

Answer:

$\displaystyle \text{Given } A(0,2) \text{ and } B(0, -2) . \text{ Let } P(h,k)$ be the point such that the sum of its distances from $\displaystyle (0,2) \text{ and } (0, -2) \text{ or } 6$ .

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

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

Squaring both sides

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

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

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

Squaring once again we get

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

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

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

$\displaystyle \text{Hence locus of } (h,k) \text{ or } 9x^2 + 5y^2 - 45 = 0$

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

Answer:

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

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

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

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

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

$\displaystyle \text{Hence locus of } (h,k) \text{ or } x^2 -2x - 6y +10 = 0$

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

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

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

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

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

$\displaystyle \text{Hence locus of } (h, k) \text{ or } x^2 = 8y^2$

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

Answer:

$\displaystyle \text{Given } A (5, 3), B (3, - 2) . \text{ Let } P(h,k)$ be the point.

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

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

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

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

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

$\displaystyle \text{Hence locus of } (h, k) is 5x-2y - 37 = 0$ or $\displaystyle 5x - 2y - 1 = 0$

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

Answer:

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

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

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

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

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

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

$\displaystyle \text{Hence locus of } (h, k) \text{ or } x^2 + y^2 = 4$

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

Answer:

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

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

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

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

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

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

$\displaystyle \text{Hence locus of } ( h, k) \text{ or } 2x - 3y - 11 = 0$ or $\displaystyle 2x-3y+17 = 0$

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

Answer:

Let $\displaystyle x \text{ and } y$ axis be the two perpendicular lines.

Let $\displaystyle A(a,0) \text{ and } B ( 0, b)$ be the two intercepts on the axes.

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

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

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

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

Now $\displaystyle AB = l$

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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