Class 11: Cartesian System of Rectangular Co-ordinates

$\displaystyle \textbf{Question 1: }\text{Find the locus of a point equidistant from the point }(2,4)\text{ and the }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(h,k)\text{ be the required point.}$
$\displaystyle \text{The distance of }P(h,k)\text{ from the }y\text{-axis is }h.$
$\displaystyle \text{The distance of }P(h,k)\text{ from }(2,4)\text{ is }\sqrt{(h-2)^2+(k-4)^2}.$
$\displaystyle \therefore h=\sqrt{(h-2)^2+(k-4)^2}$
$\displaystyle \Rightarrow h^2=(h-2)^2+(k-4)^2$
$\displaystyle \Rightarrow h^2=h^2-4h+4+k^2-8k+16$
$\displaystyle \Rightarrow k^2-4h-8k+20=0$
$\displaystyle \therefore \text{The locus of }P(h,k)\text{ is }k^2-4h-8k+20=0.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the locus is }y^2-4x-8y+20=0.$
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$\displaystyle \textbf{Question 2: }\text{Find the equation of the locus of a point which moves such that the ratio of its}$
$\displaystyle \text{distances from }(2,0)\text{ and }(1,3)\text{ is }5:4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(2,0),\ B(1,3)\text{ and let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{Given, }\frac{PA}{PB}=\frac{5}{4}$
$\displaystyle \therefore \frac{\sqrt{(h-2)^2+k^2}}{\sqrt{(h-1)^2+(k-3)^2}}=\frac{5}{4}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle \frac{(h-2)^2+k^2}{(h-1)^2+(k-3)^2}=\frac{25}{16}$
$\displaystyle \Rightarrow 16\left[(h-2)^2+k^2\right]=25\left[(h-1)^2+(k-3)^2\right]$
$\displaystyle \Rightarrow 16(h^2-4h+4+k^2)$
$\displaystyle =25(h^2-2h+1+k^2-6k+9)$
$\displaystyle \Rightarrow 16h^2-64h+64+16k^2$
$\displaystyle =25h^2-50h+25k^2-150k+250$
$\displaystyle \Rightarrow 9h^2+9k^2+14h-150k+186=0$
$\displaystyle \therefore \text{The locus of }P(h,k)\text{ is }9h^2+9k^2+14h-150k+186=0.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle 9x^2+9y^2+14x-150y+186=0.$
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$\displaystyle \textbf{Question 3: }\text{A point moves such that the difference of its distances from }(ae,0)\text{ and}$
$\displaystyle (-ae,0)\text{ is }2a.\text{ Prove that the equation of its locus is}$
$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ where }b^2=a^2(e^2-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(ae,0),\ B(-ae,0)\text{ and let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{Taking the difference of the distances as }PB-PA=2a,$
$\displaystyle \sqrt{(h+ae)^2+k^2}-\sqrt{(h-ae)^2+k^2}=2a$
$\displaystyle \Rightarrow \sqrt{(h+ae)^2+k^2}=2a+\sqrt{(h-ae)^2+k^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle (h+ae)^2+k^2=4a^2+(h-ae)^2+k^2$
$\displaystyle \qquad+4a\sqrt{(h-ae)^2+k^2}$
$\displaystyle \Rightarrow 4aeh=4a^2+4a\sqrt{(h-ae)^2+k^2}$
$\displaystyle \Rightarrow eh-a=\sqrt{(h-ae)^2+k^2}$
$\displaystyle \text{Squaring again,}$
$\displaystyle (eh-a)^2=(h-ae)^2+k^2$
$\displaystyle \Rightarrow e^2h^2-2aeh+a^2=h^2-2aeh+a^2e^2+k^2$
$\displaystyle \Rightarrow (e^2-1)h^2=a^2(e^2-1)+k^2$
$\displaystyle \Rightarrow h^2-a^2=\frac{k^2}{e^2-1}$
$\displaystyle \Rightarrow \frac{h^2}{a^2}-\frac{k^2}{a^2(e^2-1)}=1$
$\displaystyle \text{Let }b^2=a^2(e^2-1).$
$\displaystyle \therefore \frac{h^2}{a^2}-\frac{k^2}{b^2}=1$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ where }b^2=a^2(e^2-1).$
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$\displaystyle \textbf{Question 4: }\text{Find the locus of a point such that the sum of its distances from }(0,2)\text{ and}$
$\displaystyle (0,-2)\text{ is }6.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(0,2),\ B(0,-2)\text{ and let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{Given, }PA+PB=6$
$\displaystyle \therefore \sqrt{h^2+(k-2)^2}+\sqrt{h^2+(k+2)^2}=6$
$\displaystyle \Rightarrow \sqrt{h^2+(k-2)^2}=6-\sqrt{h^2+(k+2)^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle h^2+(k-2)^2=36+h^2+(k+2)^2$
$\displaystyle \qquad-12\sqrt{h^2+(k+2)^2}$
$\displaystyle \Rightarrow h^2+k^2-4k+4$
$\displaystyle =36+h^2+k^2+4k+4-12\sqrt{h^2+(k+2)^2}$
$\displaystyle \Rightarrow 36+8k=12\sqrt{h^2+(k+2)^2}$
$\displaystyle \Rightarrow 9+2k=3\sqrt{h^2+(k+2)^2}$
$\displaystyle \text{Squaring again,}$
$\displaystyle (9+2k)^2=9\left[h^2+(k+2)^2\right]$
$\displaystyle \Rightarrow 81+36k+4k^2=9h^2+9k^2+36k+36$
$\displaystyle \Rightarrow 9h^2+5k^2-45=0$
$\displaystyle \therefore \text{The locus of }P(h,k)\text{ is }9h^2+5k^2-45=0.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle 9x^2+5y^2-45=0$
$\displaystyle \text{or }\frac{x^2}{5}+\frac{y^2}{9}=1.$
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$\displaystyle \textbf{Question 5: }\text{Find the locus of a point which is equidistant from }(1,3)\text{ and the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(1,3)\text{ and let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{The distance of }P(h,k)\text{ from the }x\text{-axis is }|k|.$
$\displaystyle \text{Since }P\text{ is equidistant from }A\text{ and the }x\text{-axis,}$
$\displaystyle AP=|k|$
$\displaystyle \Rightarrow AP^2=k^2$
$\displaystyle \Rightarrow (h-1)^2+(k-3)^2=k^2$
$\displaystyle \Rightarrow h^2-2h+1+k^2-6k+9=k^2$
$\displaystyle \Rightarrow h^2-2h-6k+10=0$
$\displaystyle \therefore \text{The locus of }P(h,k)\text{ is }h^2-2h-6k+10=0.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle x^2-2x-6y+10=0.$
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$\displaystyle \textbf{Question 6: }\text{Find the locus of a point which moves such that its distance from the origin is}$
$\displaystyle \text{three times its distance from the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(h,k)\text{ be the moving point and let }O(0,0)\text{ be the origin.}$
$\displaystyle \text{The distance of }P(h,k)\text{ from the }x\text{-axis is }|k|.$
$\displaystyle \text{Given, }OP=3|k|$
$\displaystyle \Rightarrow OP^2=9k^2$
$\displaystyle \Rightarrow (h-0)^2+(k-0)^2=9k^2$
$\displaystyle \Rightarrow h^2+k^2=9k^2$
$\displaystyle \Rightarrow h^2=8k^2$
$\displaystyle \therefore \text{The locus of }P(h,k)\text{ is }h^2=8k^2.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle x^2=8y^2.$
$\displaystyle \text{Equivalently, }x=\pm2\sqrt{2}\,y.$
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$\displaystyle \textbf{Question 7: }A(5,3)\text{ and }B(3,-2)\text{ are two fixed points. Find the equation of the locus}$
$\displaystyle \text{of a point }P\text{ which moves such that the area of }\triangle PAB\text{ is }9\text{ square units.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{The area of a triangle with vertices }(x_1,y_1),\ (x_2,y_2)\text{ and }(x_3,y_3)\text{ is}$
$\displaystyle \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$
$\displaystyle \therefore \text{Area of }\triangle ABP=\frac{1}{2}\left|5(-2-k)+3(k-3)+h(3+2)\right|$
$\displaystyle \Rightarrow 9=\frac{1}{2}|-10-5k+3k-9+5h|$
$\displaystyle \Rightarrow |5h-2k-19|=18$
$\displaystyle \therefore 5h-2k-19=18\quad\text{or}\quad 5h-2k-19=-18$
$\displaystyle \Rightarrow 5h-2k-37=0\quad\text{or}\quad 5h-2k-1=0$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle 5x-2y-37=0\quad\text{or}\quad 5x-2y-1=0.$
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$\displaystyle \textbf{Question 8: }\text{Find the locus of a point such that the line segment joining }(2,0)\text{ and }(-2,0)$
$\displaystyle \text{subtends a right angle at that point.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(2,0),\ B(-2,0)\text{ and let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{Given, }\angle APB=90^\circ.$
$\displaystyle \therefore AB^2=AP^2+BP^2$
$\displaystyle \Rightarrow (2+2)^2=(h-2)^2+k^2+(h+2)^2+k^2$
$\displaystyle \Rightarrow 16=h^2-4h+4+k^2+h^2+4h+4+k^2$
$\displaystyle \Rightarrow 16=2h^2+2k^2+8$
$\displaystyle \Rightarrow h^2+k^2=4$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle x^2+y^2=4,$
$\displaystyle \text{excluding the points }(2,0)\text{ and }(-2,0).$
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$\displaystyle \textbf{Question 9: }\text{If }A(-1,1)\text{ and }B(2,3)\text{ are two fixed points, find the locus of a point }P$
$\displaystyle \text{such that the area of }\triangle PAB\text{ is }8\text{ square units.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(h,k)\text{ be the moving point.}$
$\displaystyle \text{The area of a triangle with vertices }(x_1,y_1),\ (x_2,y_2)\text{ and }(x_3,y_3)\text{ is}$
$\displaystyle \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$
$\displaystyle \therefore \text{Area of }\triangle ABP$
$\displaystyle =\frac{1}{2}\left|-1(3-k)+2(k-1)+h(1-3)\right|$
$\displaystyle =\frac{1}{2}\left|-3+k+2k-2-2h\right|$
$\displaystyle =\frac{1}{2}|-2h+3k-5|$
$\displaystyle =\frac{1}{2}|2h-3k+5|$
$\displaystyle \text{Given, }\frac{1}{2}|2h-3k+5|=8$
$\displaystyle \Rightarrow |2h-3k+5|=16$
$\displaystyle \therefore 2h-3k+5=16\quad\text{or}\quad 2h-3k+5=-16$
$\displaystyle \Rightarrow 2h-3k-11=0\quad\text{or}\quad 2h-3k+21=0$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle 2x-3y-11=0\quad\text{or}\quad 2x-3y+21=0.$
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$\displaystyle \textbf{Question 10: }\text{A rod of length }l\text{ slides between two perpendicular lines. Find the locus}$
$\displaystyle \text{of the point on the rod which divides it in the ratio }1:2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the }x\text{-axis and }y\text{-axis be the two perpendicular lines.}$
$\displaystyle \text{Let }A(a,0)\text{ and }B(0,b)\text{ be the ends of the rod on the axes.}$
$\displaystyle \text{Let }P(h,k)\text{ divide }AB\text{ internally in the ratio }AP:PB=1:2.$
$\displaystyle \text{By the section formula,}$
$\displaystyle (h,k)=\left(\frac{1(0)+2a}{1+2},\frac{1(b)+2(0)}{1+2}\right)$
$\displaystyle =\left(\frac{2a}{3},\frac{b}{3}\right)$
$\displaystyle \therefore a=\frac{3h}{2}\quad\text{and}\quad b=3k$
$\displaystyle \text{Since the length of the rod is }l,$
$\displaystyle AB=l$
$\displaystyle \Rightarrow \sqrt{a^2+b^2}=l$
$\displaystyle \Rightarrow a^2+b^2=l^2$
$\displaystyle \Rightarrow \left(\frac{3h}{2}\right)^2+(3k)^2=l^2$
$\displaystyle \Rightarrow \frac{9h^2}{4}+9k^2=l^2$
$\displaystyle \Rightarrow \frac{h^2}{4}+k^2=\frac{l^2}{9}$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle \frac{x^2}{4}+y^2=\frac{l^2}{9}.$
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$\displaystyle \textbf{Question 11: }\text{Find the locus of the mid-point of the portion of the line}$
$\displaystyle x\cos\alpha+y\sin\alpha=p\text{ which is intercepted between the coordinate axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is }x\cos\alpha+y\sin\alpha=p.$
$\displaystyle \text{When }y=0,\quad x=\frac{p}{\cos\alpha}=p\sec\alpha.$
$\displaystyle \therefore A(p\sec\alpha,0).$
$\displaystyle \text{When }x=0,\quad y=\frac{p}{\sin\alpha}=p\,\mathrm{cosec}\alpha.$
$\displaystyle \therefore B(0,p\,\mathrm{cosec}\alpha).$
$\displaystyle \text{Let }P(h,k)\text{ be the mid-point of }AB.$
$\displaystyle \therefore h=\frac{p\sec\alpha+0}{2}=\frac{p\sec\alpha}{2}$
$\displaystyle \text{and }k=\frac{0+p\,\mathrm{cosec}\alpha}{2}=\frac{p\,\mathrm{cosec}\alpha}{2}.$
$\displaystyle \Rightarrow \sec\alpha=\frac{2h}{p}\quad\text{and}\quad \mathrm{cosec}\alpha=\frac{2k}{p}$
$\displaystyle \Rightarrow \cos\alpha=\frac{p}{2h}\quad\text{and}\quad \sin\alpha=\frac{p}{2k}.$
$\displaystyle \text{Using }\cos^2\alpha+\sin^2\alpha=1,$
$\displaystyle \left(\frac{p}{2h}\right)^2+\left(\frac{p}{2k}\right)^2=1$
$\displaystyle \Rightarrow \frac{p^2}{4h^2}+\frac{p^2}{4k^2}=1$
$\displaystyle \Rightarrow \frac{1}{h^2}+\frac{1}{k^2}=\frac{4}{p^2}.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle \frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}.$
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$\displaystyle \textbf{Question 12: }\text{If }O\text{ is the origin and }Q\text{ is a variable point on }y^2=x,$
$\displaystyle \text{find the locus of the mid-point of }OQ.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }Q(a,b)\text{ be a point on }y^2=x.$
$\displaystyle \therefore b^2=a.\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle \text{Let }P(h,k)\text{ be the mid-point of }OQ.$
$\displaystyle \therefore h=\frac{0+a}{2}=\frac{a}{2}\text{ and }k=\frac{0+b}{2}=\frac{b}{2}$
$\displaystyle \Rightarrow a=2h\text{ and }b=2k$
$\displaystyle \text{Substituting these values in (i),}$
$\displaystyle (2k)^2=2h$
$\displaystyle \Rightarrow 4k^2=2h$
$\displaystyle \Rightarrow 2k^2=h$
$\displaystyle \therefore \text{The locus of }P(h,k)\text{ is }2k^2=h.$
$\displaystyle \text{Replacing }h\text{ by }x\text{ and }k\text{ by }y,\text{ the required locus is}$
$\displaystyle 2y^2=x.$
$\displaystyle \\$