5 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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Question 15 Marks

The base of an equilateral triangle with side 2a lies along the they-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer

It is given that ABC is an equilatral triangle. $\therefore $ AB = BC = AC = 2a Area of equilatral triangle $=\frac{\sqrt{3}}{4}\text{(side)}^2$ $=\frac{\sqrt{3}}{4}\times(2\text{a})^2$ $=\frac{\sqrt{3}}{4}\times4\times\text{a}^2$ $=\sqrt{3}\text{a}^2$ But, area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}.$ $\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height}=\sqrt{3}\text{a}^2$ $\Rightarrow\frac{1}{2}\times\text{BC}\times\text{OA}=\sqrt{3}\text{a}^2$ $\Rightarrow\frac{1}{2}\times2\text{a}\times\text{OA}=\sqrt{3}\text{a}^2$ $\therefore $ coordinates of A are $(\sqrt{3}\text{a},0)\text{ or }\text{OA}(-\sqrt{3}\text{a},0)$ Clearly, the coordinates of B and C are (0, -a) and (0, a) respectively.
Hence, the vertices of the triangle are $(0,\text{a}),(0,-\text{a})\text{ and }\big(-\sqrt{3}\text{a},0\big) \text{ or }(0, \text{a}),(0,-\text{a})\text{ and }\big(\sqrt{3}\text{a},0\big).$

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Question 25 Marks

Verify that the area of the triangle with verticies (2, 3), (-3, -1) remains invariant under the translation of axes when the origin is shifted to the point (-1, 3).

Answer

Let A(2, 3), B(5, 7) and C(-3, -1) represent the vertices of the triangle. $\therefore \text{ of } \triangle\text{ ABC }=\frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|$ $=\frac{1}{2}\big|2(7+1)+5(-1-3)-3(3-7)\big|$ $=\frac{1}{2}\big|16-20+12\big|$ $=4$ Since the origin is shifted to the point (-1, 3), the vertices of the $\triangle\text{ ABC }$ will be$\text{A}'(2+1,3-3), \text{B}'(5+1,7-3)\text{ and } \text{C}'(-3+1,-1-3)$
$\text{ or } \text{A}'(3,0), \text{B}'(6,4),\text{ and }\text{C}'(-2-4)$
$\text{Now }, \text{ area }\text{of } \triangle\text{A}'\text{B}'\text{C}':$
$= \frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|$ $= \frac{1}{2}\big|3(4+4)+6(-4-0)-2(0-4)\big|$ $=4$ Hence, area of the triangle remains invariant.

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Question 35 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:
$x^2 + y^2 - 5x + 2y - 5 = 0$

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 $x^2 + y^2 - 5x + 2y - 5 = 0,$ We get
$(X + h)^2 + (Y + k)^2 - 5(X - h) - 2(Y + k) + 3 = 0$
$\Rightarrow x^2 + h^2 + 2Xh + Y^2 + k^2 + 2Yk - 5X - 5h + 2Y + 2k - 5 = 0$
$\Rightarrow x^2 + Y^2+ 2Yk + 2Y + 2Xh - 5X + h^2 + k^2 - 5h + 2k - 5 = 0$
$\Rightarrow x^2 + Y^2+ (2k + 2)Y + (2h - 5)X + h^2 + k^2 - 5h + 2k - 5 = 0$
For this equation to be free from the term of first degree, we must have
$2k + 2 = 0$ and $2h - 5 = 0$
$\Rightarrow k = - 1$ and $\text{h}=\frac{5}{2}$
Hence, the origin is shifted at the point $\Big(\frac{5}{2}, -1\Big)$.

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Question 45 Marks

Four point A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that $\frac{\triangle\text{DBC}}{\triangle\text{ABC}}=\frac{1}{2}$, find x.

Answer

$\text{A}(6,3), \text{B}(-3,5),\text{C}(4,-2),\text{D}(\text{x},\text{3x})$
$\text{or }(\triangle\text{D}\text{B}\text{C})=\frac{1}{2}\big[\text{x}_1(\text{y}_2-\text{y}_1)+\text{x}_2(\text{y}_3-\text{y}_2)\big]$
$=\frac{1}{2}\big[-3(-2-3\text{x})+4(3\text{x}-5)+\text{x}(5+2)\big]$
$=\frac{1}{2}\big[1+9\text{x}+12\text{x}-20+5\text{x}+2\text{x}\big]$
$=\frac{1}{2}\big[28\text{x}-14\big]$
$=7\big[2\text{x}-1\big]$
$\text{or }(\triangle\text{ABC})=\frac{1}{2}\big[6(5+2)-3(-2-3)+4(3-5)\big]$
$=\frac{1}{2}[42+15-8]$
$\frac{\text{or }(\triangle \text{DBC})}{\text{or } (\triangle\text{ABC})}=\frac{1}{2}$
$\frac{7(2\text{x}-1)}{\frac{49}{2}}=\frac{1}{2}$
$\frac{14(2\text{x}-1)}{49}=\frac{1}{2}$
$\frac{28\text{x}-14}{49}=\frac{1}{2}$
$56\text{x}-28=49$
$56\text{x}=28+49$
$56\text{x}=77$
$\text{x}=\frac{11}{8}$

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Question 55 Marks

A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.

Answer

Let the two perpendicular lines be the coordinate axes. Let AB be a rod of length l. Let the coordinates of A and B be (a, 0) and (0, b) respectively. As the rod AB slides, the values. Let P(h, k) be a point on the locus. Then,$\text{h}=\frac{2\times \text{a}+1\times 0}{2+1}$
$\Rightarrow \text{h}=\frac{2\text{a}}{3}$
$\Rightarrow \text{a}=\frac{3\text{h}}{2}$
$\text{and }\text{k}=\frac{2\times0+\text{b}\times1}{2+1}$
$\Rightarrow \text{k}=\frac{\text{b}}{3}$
$\Rightarrow \text{b}=3\text{k}$
From $\triangle\text{AOB},$ We have
$\text{AB}^2=\text{OA}^2 +\text{OB}^2$
$\Rightarrow \text{l}^2=\big[(\text{a}-0)^2+(0,0)^2\big]+\big[(0,0)^2+(\text{a}-0)^2,\big]$
$\Rightarrow \text{l}^2=\text{a}^2+\text{b}^2$
$\Rightarrow \text{a}^2+\text{b}^2=\text{l}^2$
$\Rightarrow \bigg(\frac{3\text{h}}{2}\bigg)^2+(3\text{k})^2=\text{l}^2$
$\Rightarrow \frac{9\text{h}^2}{4}+9\text{k}^2=\text{l}^2$
$\Rightarrow \frac{\text{h}^2}{4}+\text{k}^2=\frac{\text{l}^2}{9}$
Hence, the locus of (h, k) is $\frac{\text{x}^2}{4}+\text{y}^2=\frac{\text{l}^2}{9}$

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Question 65 Marks

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$ P(h, k)$ be the point on the locus and the coordinates of a are $(a, b).$
Then,$\text{h}=\frac{\text{a}+0}{2}\text{ and }\frac{\text{b}+0}{2}=\text{k}$
$[\because$ P is thr mid-point of Q the origino $]$
$\text{h}=\frac{\text{a}}{2}\text{ and }\text{b}=2\text{k}$
$\Rightarrow \text{a}=2\text{h}\text{ and }\text{b}=2\text{k}$
point Q lies on the $y^2 = x.$ Then,
$\text{b}^2=\text{a}$
$[\because \text{Q}:\text{(a,b)}]$
$\Rightarrow (2\text{k})^2=2\text{h}$
$[\because\text{a}=2\text{h} \text{ and } \text{b}=2\text{k}]$
$\Rightarrow 4\text{k}^2=2\text{h}$
$\Rightarrow 2\text{k}^2=\text{h}$
Hence, the locus of $(h, k)$ is $2\text{y}^2=\text{x}$

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Question 75 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:
$x^2 - 12x + 4 = 0$

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 $x^2 - 12x + 4 = 0,$
We get $(X + h)^2 - 12(X + h)^2 + 4 $
$\Rightarrow X^2 + h^2 + 2 \times h - 12X - 12h + 4 = 0 $
$\Rightarrow X^2+ (2h - 12)X + h^2 - 12h + 4 = 0$
For this equation to be free from the term of first degree,
we must have $2h - 12 = 0$
$\Rightarrow \text{h}=\frac{12}{2}$
$\Rightarrow \text{h}=6$
Hence, the origin is shifted at the point $(6, \text{k}) \text{k}\in\text{R}$

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Question 85 Marks

Verify that the area of the triangle with vertice (4, 6), (7, 10) and (1, -2) remains invariant under the translation of axes when the origin is shifted to the point (-2, 1).

Answer

Let, the co-ordinate of the vertex be A(4, 6), B(7, 10) and C(1, -2).
Now area of the $\triangle \text{ABC}$ is given by,
$\triangle= \frac{1}{2}\Big|(\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2))\Big|$
$=\frac{1}{2}\big|(4(10+2)+7(-2-6)+1(6-10))\big|$
$=\frac{1}{2}\big|(48-56-4)\big|$
$=6$
After transfroming the origin to (-2, 1), the co-ordinate of the vertex will be A(2, 7), B(5, 11) and C(-1, -1).
Now the area will be:
$\triangle_1= \frac{1}{2}\Big|(\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2))\Big|$
$=\frac{1}{2}\big|(2(11+1)+5(-1-7)-1(7-11))\big|$
$=\frac{1}{2}\big|(24-40+4)\big|$
$=6$
$\text{Here }\triangle=\triangle_1$
Hence proved.

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Question 95 Marks

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

Answer

Let P(h, k) be any point on the locus. Then,Area (PAB) = 8 sq units
$\Rightarrow \frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|=8$
$\Rightarrow \frac{1}{2}\big|-1(3-\text{k})+2(\text{k}-1)+\text{h}(1-3)\big|=8$
$\Rightarrow \frac{1}{2}\big|-3+\text{k}+2\text{k}+2-2\text{h}\big|=8$
$\Rightarrow \frac{1}{2}\big|-2\text{h}+3\text{k}-5\big|=8$
$\Rightarrow \big|-2\text{h}+3\text{k}-5\big|=16$
$\Rightarrow -2\text{h}+3\text{k}-5=\pm16$
$\Rightarrow 2\text{h}-3\text{k}+5\pm16=0$
$\Rightarrow 2\text{h}-3\text{k}+21=0 \ \text{or, }\ 2\text{h}-3\text{k}-11=0$
Hence, the locus of(h, k) is $2\text{h}-3\text{k}+21=0 \ \text{or, }\ 2\text{h}-3\text{k}-11=0$

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Question 105 Marks

If the line segment joining the points $P(x_1, y_1)$ and $Q(x_2, y_2)$ subtends an angle $\alpha$ at the origin O, prove that: $\text{O}\text{P}.\text{O}\text{Q}\cos\alpha =\text{x}_1\text{x}_2+\text{y}_1\text{y}_2. $

Answer

It is given that $O$ is the origin.
Then,
$\text{OQ}^2=\text{x}_2^2+\text{y}_2^2$
$\text{OP}^2=\text{x}_1^2+\text{y}_1^2$
and, $\text{OP}^2=(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2$
Using cosine fromula in Rectangle OPQ, we have
$\text{OP}^2=\text{OP}^2+\text{OQ}^2-2.(\text{OP})(\text{OQ})\cos\alpha$
$\Rightarrow (\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2=\text{x}_2^2+\text{y}_2^2+\text{x}_1^2+\text{y}_1^2-2(\text{OP}).(\text{OQ})\cos\alpha$
$\Rightarrow -2\text{x}_1\text{x}_2-2\text{y}_1\text{y}_2=-2\text{OP}.\text{O}\text{Q}\cos\alpha$
$\Rightarrow \text{x}_1\text{x}_2+\text{y}_1\text{y}_2=\text{OP}.\text{OQ}\cos\alpha$
$\Rightarrow\text{OP}.\text{OQ}\cos\alpha =\text{x}_1\text{x}_2+\text{y}_1\text{y}_2$
Hence, Proved.

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Question 115 Marks

The vertices of a triangle ABC are A(0, 0), B(2, -1) and C(9, 2). Find cos B.

Answer

We know that,$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{a}\text{c}}$
Where a = BC, b = CA and C = AB are the sides of the triangle ABC.
We have,
$\text{a}=\text{BC}=\sqrt{(9-2)^2+(2+1)^2}=\sqrt{49+9}=\sqrt{58}$
$\text{b}=\text{CA}=\sqrt{(0-9)^2+(0+2)^2}=\sqrt{81+4}=\sqrt{85}$
$\text{and}, \text{ c}=\text{AB}=\sqrt{(2-0)^2+(-1-0)^2}=\sqrt{4+1}=\sqrt{5}$
$\therefore\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{a}\text{c}}$
$=\frac{58+5-85}{2\times\sqrt{58}\times\sqrt{5}}$
$=\frac{63-85}{2\sqrt{290}}$
$=\frac{-22}{2\sqrt{290}}=\frac{-11}{\sqrt{290}}$
Hence, $\cos\text{B}=\frac{-11}{\sqrt{290}}$

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Question 125 Marks

Find the coording of the centre of the circle inscribed in a triangle whose vertics are (-36, 7), (20, 7) and (0, -8).

Answer

Let A(-36, 7), B(20, 7) and c(0, -8) be the vertices of the triangle ABC.Now,
$\text{a}=\text{BC}=\sqrt{(0-20)^2+(-8-7)^2}$
$=\sqrt{400+225}$
$=\sqrt{625}$
$=25$
$\text{and }\text{c }=48=\sqrt{(20+36)^2+(7-7)^2}$
$=\sqrt{(56)^2}$
$=56$
The coordinates of the center of the cricle are,
$\bigg(\frac{\text{a}\text{x}_1+\text{b}\text{x}_2+\text{c}\text{x}_3}{\text{a}+\text{b}+\text{c}},\frac{\text{a}\text{y}_1+\text{b}\text{y}_2+\text{c}\text{y}_3}{\text{a}+\text{b}+\text{c}}\bigg)$
$\text{or }, \bigg[\frac{25\times(-36)+39\times20+56\times0}{25+39+56},\frac{25\times7+39\times7+56\times(-8)}{25+39+56}\bigg]$
$\text{or},\bigg[\frac{-900+780}{120},\frac{175+273-448}{120}\bigg]$
$\text{or},\bigg[\frac{-120}{120},\frac{0}{120}\bigg]$
$\text{or},(-1,0)$
Hence, the coordinates of the centre of the crcle are (-1, 0).

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Question 135 Marks

Find the distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ when,

$PQ$ is parallel to the $y-$axis.
$PQ$ is parallel to the $x-$axis.

Answer

It is given that $P(x_1, y_1)$ and $Q(x_2, y_2)$ are two points.

$PQ$ is parallel to the $y-$axis.

$\therefore\text{ x}_1=\text{x}_2 \ . . .(1)$
$\therefore \text{PQ}=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$
$=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$
$[$using equaction $1]$
$=\bigg|\sqrt{(\text{y}_2-\text{y}_1)^2}\bigg|$
$=\big|\text{y}_2-\text{y}_1\big|$

$PQ$ is parallel to the $x-$axis.

$\therefore\text{ y}_1=\text{y}_2 \ . . .(2)$
$\therefore \text{PQ}=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$
$=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$
$[$using equaction $2]$
$=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2}\bigg|$
$\therefore \text{ PQ}=\big|\text{x}_2-\text{x}_1\big|$

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Question 145 Marks

At What point the origin be shifted so that the equation $x^2+ xy - 3x - y + 2 = 0 $ cantain any first degree term and constand term?

Answer

We have$,x^2+ xy - 3x - y + 2 = 0 ...(i)$
Let the origin be shifted to $(h, k)$. Then $x = X$ and $y = Y + k.$
Substituting x = X + h, y = Y + k in the given equation, we get
$(X + h)^2 + (X + h)(Y + k) - 3(X + h) - (Y + k) + 2 =0$
$\Rightarrow X^2 + h^2 + 2Xh + XY + Xk + Yh + hk - 3k - 3h - Y - k + 2 = 0$
$\Rightarrow X^2+ XY + 2 \times Xk + Yh - Y3 - X + h^2 + hk - 3h - k + 2 = 0$
$\Rightarrow X^2 + (2Xh + Xk - 3X) + XY + (Yh-Y) + (h^2 + hk - 3h - k + 2) = 0$
$\Rightarrow X^2 + (2h + k - 3)X + XY + (h - 1)Y + (h^2 + hk - 3h - k + 2) = 0$
From this equation to be free from first degree and the constant term, we must have,
$2h + k - 3 = 0......(ii)$
$h - 1 = 0$
$\Rightarrow h = 1......(iii)$
And $h^2 + hk - 3h - k + 2 = 0.....(iv)$
Putting $h = 1$ in equation (ii), we get
$2 + k - 3 = 0$
$\Rightarrow k = 1$
Putting $h = 1$ and $k = 1$ in equation $(iv),$ we get
$(1)^2+ 1 - 3 - 1 + 2 = 0.$
Hence, the value of h and k satisfies the equation $(iv)$
$\therefore$ The origin is shifted at the point $(1, 1).$

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Question 155 Marks

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

Answer

Let P(h, k) be any point on the locus and let A(ae, 0) and B(-ae, 0) be the given points. By the given condition$\text{PA}-\text{PB}=2\text{a}$
$\Rightarrow \text{PA}=2\text{a}+\text{PB}$
$\Rightarrow \sqrt{(\text{ae}-\text{h})^2+(0-\text{k})^2}=2\text{a}+\sqrt{(-\text{ae}-\text{h})^2+(0-\text{k})^2}$
$\Rightarrow (\text{ae}-\text{h})^2+\text{k}^2= \big(2\text{a}+\sqrt{(\text{ae}-\text{h}^2)+\text{k}^2}$ [Taking square on both sides]
$\Rightarrow \text{(ae)}^2+\text{h}^2-2\text{aeh}+\text{k}^2 =4\text{a}^2+(\text{a}\text{e}+\text{h})^2\\\ \ \ +\text{k}^2+2\times2\text{a}\times\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow \text{h}^2+\text{k}^2+\text{(ae)}^2-2\text{aeh} =4\text{a}^2+(\text{a}\text{e})^2+\text{h}^2\\\ \ \ +2\text{h}\text{a}\text{e}+\text{k}^2+4\text{a}\times\sqrt{(\text{a}\text{e}^2+\text{h}^2)+\text{k}^2}$
$\Rightarrow -4\text{a}-2\text{a}\text{e}\text{h}-2\text{a}\text{e}\text{h}=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -4\text{a}^2-4\text{a}\text{e}\text{h}=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -4\big[\text{a}^2+\text{a}\text{e}\text{h}\big]=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -\big[\text{a}^2+\text{a}\text{e}\text{h}\big]=\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -\text{a}[\text{a}+\text{e}\text{h}\big]=\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$ $\Rightarrow -[\text{a}+\text{e}\text{h}\big]=\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$ $\Rightarrow (\text{a}+\text{e}\text{h})^2=\bigg(\sqrt{(\text{a}\text{e}+\text{h}^2)+\text{k}^2}\bigg)^2$ [Taking square on both sides] $\Rightarrow \text{a}^2+(\text{e}\text{h})^22\text{h}\text{a}\text{e}=(\text{a}\text{e}+\text{h})^2+\text{k}^2$ $\Rightarrow \text{a}^2+(\text{e}\text{h})^2+2\text{h}\text{a}\text{e}=(\text{a}\text{e})+\text{h}^2+2\text{h}\text{a}\text{e}+\text{k}^2$ $\Rightarrow \text{a}^2+\text{e}^2\text{h}^2=\text{a}^2\text{e}^2+\text{h}^2+\text{k}^2$ $\Rightarrow \text{e}^2\text{h}^2-\text{h}^2-\text{k}^2=\text{a}^2\text{e}^2-\text{a}^2$ $\Rightarrow \text{h}^2(\text{e}^2-1)-\text{k}^2=\text{a}^2(\text{e}^2-1)$ $\Rightarrow \frac{\text{h}^2(\text{e}^2-1)}{\text{a}^2(\text{e}^2-1)}-\frac{\text{k}^2}{\text{a}^2(\text{e}^2-1)}=1$ $\Rightarrow \frac{\text{h}^2}{\text{a}^2}-\frac{\text{k}^2}{\text{b}^2},\text{Where }\text{b}^2=\text{a}^2(\text{e}^2-1)$ $\therefore$ The locus of (h, k) is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1 $ Hence proved.

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Question 165 Marks

Find a point on the x-axis, which is equidistant from the point (7, 6) and (3, 4).

Answer

It is given that Clie on the x-axis. Let coordinates of C be (x, 0).
Now, C is equidistant from the point A(7, 6) and B(3, 4).
$\therefore \text{ AB}=\text{BC}\ [\text{given}]$
$\Rightarrow \text{AC}^2=\text{BC}^2$
$\Rightarrow \bigg[\sqrt{(\text{x}-7)^2+(0-6)^2}\bigg]= \bigg[\sqrt{(\text{x}-3)^2+(0-4)^2}\bigg]$
$\Rightarrow (\text{x}-7)^2+(-6)^2=(\text{x}-3)^2+(-4)^2$
$\Rightarrow \text{x}^2+49-14\text{x}+36=\text{x}^2+9-6\text{x}+16$
$\Rightarrow 49+36-36-16-9=\text{x}^2-\text{x}^2-6\text{x}+14\text{x}$
$\Rightarrow 85-25=8\text{x}$
$\Rightarrow 60=8\text{x}$
$\Rightarrow 8\text{x}=60$
$\Rightarrow \text{x}=\frac{60}{8}=\frac{15}{2}$
Hence, coordinates of c are $\bigg(\frac{15}{2},0\bigg).$

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5 Marks Questions - MATHS STD 11 Science Questions - Vidyadip