(Each question 3 marks)

Question 13 Marks

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

Answer

Let P(h, k) be any point on the locus let A(0, 2) and B(0, -2) be the given point. By the given condition $\text{PA}+\text{PB}=6$ $\Rightarrow \sqrt{(\text{h}-0)^2+(\text{k}-2)^2}+\sqrt{(\text{h}-0)^2+(\text{k}+2)^2}=6$ $\Rightarrow \sqrt{\text{h}+(\text{k}-2)\text{h}^2}=6-\sqrt{\text{h}^2+(\text{k}+2)^2}$ $\Rightarrow \text{h}^2+(\text{k}-2)\text{h}^2=36-12\sqrt{\text{h}^2+(\text{k}+2)^2}+\text{h}^2+(\text{k}+2)^2$ $\Rightarrow 8\text{k}-36=-12\sqrt{\text{h}^2+(\text{k}+2)^2}$ $\Rightarrow (2\text{k}+9)=3\sqrt{\text{h}^2+(\text{k}+2)^2}$ $\Rightarrow (2\text{k}+9)^2=9\Big(\text{h}^2+(\text{k}+2)^2\Big)$ $\Rightarrow 4\text{k}^2+36\text{k}+81=9\text{h}^2+9\text{k}^2+36\text{k}+36$$\Rightarrow 9\text{h}^2+5\text{k}^2=45$
Hence, locus of (h, k) is $ 9\text{h}^2+5\text{k}^2=45.$

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Question 23 Marks

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

Answer

Let P(h, k) be any point on the locus and let A(2, 0) and B(1, 3). Then, $\frac{\text{PA}}{\text{BP}}=\frac{5}{4}$ $\frac{\text{PA}^2}{\text{BP}^2}=\frac{25}{16}$$\Rightarrow \frac{\bigg[\sqrt{(\text{h}-2)^2+(\text{k}-0)^2}\bigg]}{\bigg[\sqrt{(\text{h}-1)^2+(\text{k}-0)^3}\bigg]}=\frac{25}{16}$
$\Rightarrow \frac{(\text{h}-2)^2+\text{k}^2}{(\text{h}-1)^2+(\text{k}^3-3)^2}=\frac{25}{16}$
$\Rightarrow \frac{\text{h}^2+4-4\text{h}+\text{k}^2}{\text{h}^2+1-2\text{h}+\text{k}^2+9-6\text{k}}=\frac{25}{16}$
$\Rightarrow \frac{(\text{h}^2-4\text{h}+\text{k}^2+4)}{\text{h}^2+\text{k}^2-2\text{h}-6\text{k}+10}=\frac{25}{16}$
$\Rightarrow 16(\text{h}^2-4\text{h}+\text{k}^2+4)=25(\text{h}^2+\text{k}^2-2\text{h}-6\text{k}+10)$
$\Rightarrow 16\text{h}^2-64\text{h}+16\text{k}^2+64=25\text{h}^2+25\text{k}^2-50\text{h}-150\text{k}+250$
$\Rightarrow 25\text{h}^2-16\text{h}^2+25\text{k}^2-16\text{k}^2-50\text{h}+64\text{h}-150\text{k}+250-65=0$
$\Rightarrow 9\text{h}^2+9\text{k}^2+14\text{k}-150\text{k}+186=0$ Hence, locus of (h, k) is $ 9\text{h}^2+9\text{k}^2+14\text{k}-150\text{k}+186=0$

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Question 33 Marks

What does the equation $(\text{a} − \text{b}) (\text{x}^2 + \text{y}^2) −2\text{abx} = 0$ become if the origin is shifted to the point $\Bigg(\frac{\text{a}\text{b}}{\text{a}-\text{b}},0\Bigg)$ without rotation?

Answer

We have,$(\text{a} − \text{b}) (\text{x}^2 + \text{y}^2) −2\text{abx} = 0$
Substituting $\text{x}=\text{x}+\frac{\text{ab}}{\text{a-b}},\text{y}=\text{Y}$ in the given equation, we get
$(\text{a}-\text{b})\Bigg[\Big(\text{x}+\frac{\text{ab}}{\text{a-b}}\Big)+\text{Y}^2\Bigg]-2\text{ab}\Bigg[\text{x}+\frac{\text{ab}}{\text{a-b}}\Bigg]=0$
$\Rightarrow (\text{a}-\text{b})\Bigg[\text{x}^2+\Big(\frac{\text{ab}}{\text{ab}}\Big)^2+2\frac{\text{xab}}{\text{a-b}}+\text{Y}^2\Bigg]-2\text{ab}\text{x}-2\frac{\text{(ab)}^2}{\text{a-b}}=0$
$\Rightarrow (\text{a-b})\Bigg[\frac{\text{x}^2(\text{a-b})^2+(\text{ab})\text{h}^2+2\text{xab}(\text{a-b})+\text{Y}^2(\text{a-b})^2}{(\text{a-b})^2}\Bigg],\frac{2\text{abx}(\text{a-b})+2(\text{ab})^2}{\text{a-b}}=0$
$\Rightarrow\frac{\text{x}^2(\text{a-b})^2+(\text{ab})^2+2\text{ab}(\text{a-b})+\text{Y}^2(\text{a-b})^2}{\text{a-b}},\frac{2\text{ab}(\text{a-b})+2(\text{ab})^2}{\text{a-b}}=0$
$\Rightarrow\text{x}^2(\text{a}-\text{b})^2+\text{Y}^2(\text{a}-\text{b})^2+(\text{ab})^2+2\text{ab}(\text{a-b})=2\text{ab}(\text{a-b})+2(\text{ab})^2$
$\Rightarrow (\text{a-b})^2(\text{x}^2+\text{Y}^2)=(\text{a}\text{b})^2$
$\Rightarrow (\text{a-b})^2(\text{x}^2+\text{Y}^2)=\text{a}^2\text{b}^2$

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Question 43 Marks

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

Answer

Let P(h, k) be any point on the locus and let O(0, 0) be the origin.By the given condition
$\text{OP}=3\text{k}$ $[ \therefore$ k is the diffance of point from x-axis $]$
$\Rightarrow \text{OP}^2=9\text{k}^2$
$\Rightarrow \Big(\sqrt{(0-\text{h})^2+(0-\text{k})^2}\Big)^2=9\text{k}$
$\Rightarrow \text{h}^2+\text{k}^2=9\text{k}^2$
$\Rightarrow \text{h}^2=9\text{k}^2-\text{k}^2$
$\Rightarrow \text{h}^2=8\text{k}^2$
Hence, locus of (h, k) is $\text{x}^2=8\text{y}^2$

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Question 53 Marks

Find what the following equations become when the origin is shifted to the point $(1, 1). x^2 + xy − 3y^2 − y + 2 = 0$

Answer

We have, $x^2+x y-3 y^2-y+2=0$.....(i) Substituting $x=x+1, y=y+1$ in the equcation(i), we get $\Rightarrow(X+1)^2+(X+1)$$
(Y+1)-3(Y+1)^2-(Y+1)+2=0$
$\Rightarrow X^2+1+2 X+X Y+X+Y+1-3\left(Y^2+1+2 Y\right)-Y-1+2=0$
$\Rightarrow X^2+X Y+3 X+3-3 Y^2-3-6 Y=0$

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Question 63 Marks

Find what the following equations become when the origin is shifted to the point $(1, 1).x^2 - y^2 - 2x + 2y = 0$

Answer

We have, $x^2-y^2-2 x+2 y=0 . . . .(i)$
Substituting $x=x+1, y=y+1$ in the equcation(i), we get $\Rightarrow(X+1)^2+(Y+1)^2-2(X+1)+2(Y+1)=0 \Rightarrow X^2+1+$
$2 X-\left(Y^2+1+2 Y\right)-2 X-2+2 Y+2=0$
$\Rightarrow X^2+1-Y^2-1-2 Y+2 X=0$
$\Rightarrow X^2+2 X+1-\left(Y^2+2 Y+1\right)$
$\Rightarrow(X+1)^2-(Y+1)^2$
$\Rightarrow X^2-Y^2=0$

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Question 73 Marks

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

Let P(h, k) be any point on the locus. Then,Area (PAB) = 9 sq units
$\Rightarrow \frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|=9$
$\Rightarrow \big|5(-2-\text{k})+3(\text{k}-3)+\text{h}(3+12)\big|=18$
$\Rightarrow \big|-10-5\text{k}+3\text{k}-9\text{k}+15\big|=18$
$\Rightarrow \big|5\text{h}-2\text{k}-19\big|=18$
$\Rightarrow 5\text{h}-2\text{k}-19=\pm18$
$\Rightarrow 5\text{h}-2\text{k}-19 \mp18=0$
$\Rightarrow 5\text{h}-2\text{k}-37=0 \ \text{or, }\ 5\text{h}-2\text{k}-1=0$
Hence, the locus of(h, k) is $5\text{h}-2\text{k}-37=0 \ \text{or, }\ 5\text{h}-2\text{k}-1=0$

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Question 83 Marks

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first deree terms:$Y^2 + X^2 - 4X - 8Y + 3 = 0$

Answer

Let the origin be shifted to $(h, k)$. Then, $x=X+h$ and $y=Y+k$. Substituting $x=X+h, y=Y+k$ in the equation $Y^2$
$+\mathrm{X}^2-4 \mathrm{X}-8 \mathrm{Y}+3=0$, We get $(\mathrm{Y}+\mathrm{k})^2+(\mathrm{X}+\mathrm{h})^2-4(\mathrm{X}-\mathrm{h})-8(\mathrm{Y}+\mathrm{k})+3=0 \Rightarrow \mathrm{Y}^2+\mathrm{k}^2+2 \mathrm{Yk}+\mathrm{X}^2+\mathrm{h}^2+2 \mathrm{Xh}-4 \mathrm{X}-$
$4 \mathrm{~h}-8 \mathrm{Y}-8 \mathrm{k}+3=0 \Rightarrow \mathrm{Y}^2+\mathrm{X}^2+2 \mathrm{Yk}-8 \mathrm{Y}+2 \mathrm{Xh}-4 \mathrm{X}+\mathrm{k}^2+\mathrm{h}^2-4 \mathrm{~h}-8 \mathrm{k}+3=0 \Rightarrow \mathrm{Y}^2+\mathrm{X}^2+(2 \mathrm{k}-8) \mathrm{Y}+(2 \mathrm{~h}-4) \mathrm{X}+$
$\left(k^2+h^2-4 h-8 k+3\right)=0$ For this equation to be free from the term of first degree, we must have $2 k-8=0$ and $2 h$
$-4=0 \Rightarrow k=4$ and $h=2$ Hence, the origin is shifted at the point $(2,4)$.

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Question 93 Marks

Find what the following equations become when the origin is shifted to the point $(1, 1)? x^2 + xy − 3x − y + 2 = 0$

Answer

We have $x^2+x y-3 x-y+2=0$
Substituting $x=X+1, y+1$ in the given equation, we get
$(X+1)^2+(X+1)(Y+1)-3(X+1)-(Y+1)+2=0$
$\Rightarrow x^2+1+2 X+X Y+X+Y+1-3 X-3-Y-1+2=0$
$\Rightarrow x^2+X Y=0$

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Question 103 Marks

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

Let P(h, k) be the variable and let A(2, 0) and B(-2, 0) be the given point.Then $\angle\text{APB}=\frac{\pi}{2}$
$\Rightarrow\text{A}\text{B}^2=\text{P}\text{B}^2+\text{P}\text{B}^2$
$\Rightarrow (2+2)^2+0=(2-\text{h})^2+(0-\text{k})^2+(-2-\text{h})^2+(0-\text{k})^2$
$\Rightarrow 16=4+\text{h}^2-4\text{h}+\text{k}^2+4+\text{h}^2+4\text{h}+\text{k}$
$\Rightarrow 16=2\text{h}^2+2\text{k}^2+8$
$\Rightarrow 2\text{h}^2+2\text{k}^2+8-16=0$
$\Rightarrow 2\text{h}^2+2\text{k}^2-8=0$
$\Rightarrow \text{h}^2+\text{k}^2-4=0$
Hence, the locus of (h, k) is $\text{h}^2+\text{k}^2-4=0$

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Question 113 Marks

Find what the following equations become when the origin is shifted to the point (1, 1).xy - x - y + 1 = 0

Answer

We have,xy - x - y + 1 = 0.....(i)
Substituting x = x + 1, y = y + 1 in the equcation(i), we get⇒ (X + 1)(Y + 1) - (X + 1) - (Y + 1) + 1 = 0
⇒ XY + X + Y + 1 - X - 1 - Y - 1 + 1 = 0
⇒ XY = 0

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Question 123 Marks

Find what the following equations become when the origin is shifted to the point $(1, 1). xy − y^2 − x + y = 0$

Answer

We have, $x y-y^2-x+y=0 \ldots \ldots$.(i) Substituting $x=x+1, y=y+1$ in the equcation(i), we get $(X+1)(Y+1)-(Y+1)^2$
$(X+1)+(Y+1)=0$
$\Rightarrow X Y+X+Y+1-\left(Y^2+1+2 Y\right)-X-1+Y+1=0$
$\Rightarrow X Y+2 Y+1-Y^2-1-2 Y=0$
$\Rightarrow X Y-Y^2=0$

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Question 133 Marks

Find the locus of the mid-point of the portion of the line x cos α + y sin α = p which is intercepted between the axes.

Answer

Given, line is $\text{x}\cos\alpha+\text{y}\sin\alpha=\text{P}$
$\frac{\text{x}}{\frac{\text{p}}{\cos\alpha}}+\frac{\text{y}}{\frac{\text{p}}{\sin\alpha}}=1$
Intercepts on x axis is $\frac{\text{P}}{\cos\alpha}\text{ and }\text{y}-\text{axis }\text{is } \frac{\text{P}}{\sin\alpha}$
Let, P(x, y) be the mid-point of AB.
$\text{(x,y)}=\Bigg[\frac{\frac{\text{P}}{\cos\alpha}+0}{2},\frac{\frac{\text{P}}{\sin\alpha}+0}{2}\Bigg]$
$\therefore \text{ x}={\frac{\text{P}}{2\cos\alpha}}, \text{ y}={\frac{\text{P}}{2\sin\alpha}}$
$2\cos\alpha=\frac{\text{P}}{\text{x}}, 2\sin\alpha=\frac{\text{P}}{\text{y}}$
Square both sides,
$4\cos^2\alpha=\frac{\text{P}^2}{\text{x}^2}......(1)$
$\text{And}$
$4\sin^2\alpha=\frac{\text{P}^2}{\text{y}^2}......(2)$
$[(1)+(2)]$
$4\cos^2\alpha+4\sin^2\alpha=\frac{\text{P}^2}{\text{x}^2}+\frac{\text{P}^2}{\text{y}^2}$
$4=\frac{\text{P}^2(\text{x}^2+\text{y}^2)}{\text{x}^2\text{y}^2}$
$4\text{x}^2\text{y}^2=\text{P}^2(\text{x}^2+\text{y}^2)$

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