Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blank.
The equation of the circle circumscribing the triangle whose sides are the lines $y = x + 2, 3y = 4x, 2y = 3x$ is ___________.

Answer

The equation of the circle circumscribing the triangle whose sides are the lines $y = x + 2, 3y = 4x, 2y = 3x$ is 0.
Solution:
Given equation of line are, $y=x+2 \ldots$...(i) $3 y=4 x$.....(ii) $2 y=3 x \ldots$...(iii)
Solving these line, we get points of intersection $A(6,8), B(4,6)$ and $C(0,0)$
Let the equation of circle circumscribing the given triangle be $x^2+y^2+2 g x+2 f y+c=0$ Since the point $A(6,9), B(4$,
6 ) and $C(0,0)$ lie on this circle,
we have $36+64+12 g+16 f+c=0$
$\Rightarrow 12 g+16 f + c =-100$ Also, $16+36+8 g+12 f + c =0$
$\Rightarrow 8 g+12 f + c =-52$ And $C =0$ Putting $c =0$ in eqs. (iv) and (v)
we get $3 g+4 f=-25$ and $2 g+3 f=-13$ On solving these,
we get g $= -23 and f = 11$
So, the equation of circle is, $x^2 + y^2 - 46x + 22y + 0 = 0$
$ \Rightarrow x^2 + y^2 - 46x + 22y = 0$

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Question 21 Mark

Fill in the blank.
The equation of the ellipse having foci (0, 1), (0, -1) and minor axis of length 1 is _____________.

Answer

The equation of the ellipse having foci (0, 1), (0, -1) and minor axis of length 1 is 1.
We know that the foci of the ellipse are $(0,\pm\text{ae})$ and given foci are $(0,\pm1),$ so ae = 1
Length of minor axis = 2b = 1
$\Rightarrow\text{b}=\frac{1}{2}$
We know that $\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\Big(\frac{1}{2}\Big)^2=\text{a}^2-\text{a}^2\text{e}^2$
$\Rightarrow\frac{1}{4}=\text{a}^2-1$
$\Rightarrow\text{a}^2=1+\frac{1}{4}=\frac{5}{4}$
$\therefore$ Equation of ellipse is,
$\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{a}^2}=1$
$\Rightarrow\frac{\text{x}^2}{\frac{1}{4}}+\frac{\text{y}^2}{\frac{5}{4}}=1$
$\Rightarrow\frac{4\text{x}^2}{1}+\frac{4\text{y}^2}{5}=1$
Hence, the value of the filler is $\frac{4\text{x}^2}{1}+\frac{4\text{y}^2}{5}=1$

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Question 31 Mark

Fill in the blank.
The equation of the parabola having focus at $(-1, -2)$ and the directrix $x - 2y + 3 = 0$ is _____________.

Answer

The equation of the parabola having focus at $(-1, -2)$ and the directrix $x - 2y + 3 = 0$ is $0$.
Let $(x_1, y_1)$ be any point on the parabola. According to the definition of the parabola
$\sqrt{(\text{x}_1+1)^2+(\text{y}_1+2)^2}=\Bigg|\frac{\text{x}_1-2\text{y}_1+3}{\sqrt{(1)^2+(-2)^2}}\Bigg|$
Squaring both sides, we get
$\text{x}^2_1+1+2\text{x}_1+\text{y}^2_1+4+4\text{y}_1=\frac{\text{x}^2_1+4\text{y}^2_1+9-4\text{x}_1\text{y}_1-12\text{y}_1+6\text{x}_1}{5}$ $\Rightarrow\text{x}^2_1+\text{y}^2_1+2\text{x}_1+4\text{y}_1+5=\frac{\text{x}^2_1+4\text{y}^2_1-4\text{x}_1\text{y}_1-12\text{y}_1+6\text{x}_1+9}{5}$ $\Rightarrow5\text{x}^2_1+5\text{y}^2_1+10\text{x}_1+20\text{y}_1+25\\=\text{x}^2_1+4\text{y}^2_1-4\text{x}_1\text{y}_1-12\text{y}_1+6\text{x}_1+9$ $\Rightarrow4\text{x}^2_1+\text{y}^2_1+4\text{x}_1+32\text{y}_1+4\text{x}_1\text{y}_1+16=0$
Hence, the value of the filler is $4x^2 + y^2 + 4xy + 4x + 32y + 16 = 0$

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Question 41 Mark

Fill in the blank.
The equation of the hyperbola with vertices at $(0,\pm6)$ and eccentricity $\frac{5}{3}$ is ___________ and its foci are ___________.

Answer

The equation of the hyperbola with vertices at $(0,\pm6)$ and eccentricity $\frac{5}{3}$ is $\frac{\text{y}^2}{36}-\frac{\text{x}^2}{64}=1$ and its foci are $(0,\pm10)$Solution:
Let equation of hyperbola is $-\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
Vertices are $(0,\pm\text{b})\therefore\text{b}=6$ and $\text{e}=\frac{5}{3}$
We know that $\text{e}=\sqrt{1+\frac{\text{a}^2}{\text{b}^2}}$
$\Rightarrow\frac{5}{3}=\sqrt{1+\frac{\text{a}^2}{36}}$
$\Rightarrow\frac{25}{9}=1+\frac{\text{a}^2}{36}$
$\Rightarrow\frac{\text{a}^2}{36}=\frac{25}{9}-1=\frac{16}{9}$
$\Rightarrow\text{a}^2=\frac{16}{9}\times36$
$\Rightarrow\text{a}^2=64$
So, the equation of the hyperbola is,
$\frac{-\text{x}^2}{64}+\frac{\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{y}^2}{36}-\frac{\text{x}^2}{64}=1$
and $\text{foci}=(0,\pm\text{be})=\Big(0,\pm6\times\frac{5}{3}\Big)=(0,\pm10)$
Hence, the value of the filler is $\frac{\text{y}^2}{36}-\frac{\text{x}^2}{64}=1$ and $(0,\pm10)$

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Question 51 Mark

Fill in the blank.
An ellipse is described by using an endless string which is passed over two pins. If the axes are 6cm and 4cm, the length of the string and distance between the pins are ____________.

Answer

An ellipse is described by using an endless string which is passed over two pins. If the axes are 6cm and 4cm, the length of the string and distance between the pins are $6+2\sqrt{5}$Solution:
Let equation of the ellipse be $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ According to the question, a = 3 and b = 2 Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$ $\therefore\ \text{e}^2=2-\frac{\text{b}^2}{\text{a}^2}=1-\frac{4}{9}=\frac{5}{9}$ $\therefore\ \text{e}=\frac{\sqrt{5}}{3}$ From the definition of the ellipse for any point P on the ellipse, we have SP + S'P = 2a, where S and S' are foci $\therefore$ Length of the endless string = SP + S'P + SS' $=2\text{a}+\text{ae}=2(3)+(2)(3)\times\frac{\sqrt{5}}{3}=6+2\sqrt{5}$

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Question 61 Mark

Fill in the blank.
The equation of the circle having centre at (3, -4) and touching the line 5x + 12y - 12 = 0 is __________.
[Hint: To determine radius find the perpendicular distance from the centre of the circle to the line]

Answer

The equation of the circle having centre at (3, -4) and touching the line 5x + 12y - 12 = 0 is $=\Big(\frac{45}{13}\Big)^2$
Solution:
Given equation of the line is 5x + 12y - 12 = 0 and the centre is (3, 4)

CP = radius of the circle
$\Bigg|\frac{5\times3+12\times(-4)-12}{\sqrt{(5)^2+(12)^2}}\Bigg|=\text{r}$
$\Rightarrow\Big|\frac{15-48-12}{13}\Big|=\text{r}$
$\Rightarrow\Big|\frac{-45}{123}\Big|=\text{r}$
$\Rightarrow\text{r}^2=\frac{2025}{169}$
So, the equation of the circle is,
$(\text{x}-3)^2+(\text{y}+4)^2=\Big(\frac{45}{13}\Big)^2$
Hence, the value of the filler is $(\text{x}-3)^2+(\text{y}+4)^2=\Big(\frac{45}{13}\Big)^2$

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