3 Marks Question

Question 13 Marks

Find the coordinates of the centre and radius of the circle $(x \cos \alpha+y \sin \alpha-a)^2+$\[(x \sin \alpha-y \cos \alpha-b)^2=k^2 .\]

Answer

Equation of given circle is :
\[
\begin{array}{c}
(x \cos \alpha+y \sin \alpha-a)^2+(x \sin \alpha-y \cos \alpha \\
-b)^2=k^2 \\
(x \cos \alpha+y \sin \alpha)^2-2(x \cos \alpha+y \sin \alpha) .
\end{array}
\]
\[
\begin{array}{l}
a+a^2+(x \sin \alpha-y \cos \alpha)^2-2 \times \\
(x \sin \alpha-y \cos \alpha) \times b+b^2=k^2 \\
\Rightarrow x^2\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+y^2\left(\sin ^2 \alpha+\cos ^2 \alpha\right)+ \\
2(b \cos \alpha-a \sin \alpha) \cdot y-2(a \cos \alpha+ \\
b \sin \alpha) \cdot x+a^2+b^2-k^2=0 \\
\Rightarrow x^2+y^2-2(a \cos \alpha+b \sin \alpha) \cdot x- \\
2(a \sin \alpha-b \cos \alpha) y+a^2+b^2-k^2=0 \\
\text { Here, } \quad g=-(a \cos \alpha+b \sin \alpha), f=-(a \sin \alpha- \\
b \cos \alpha), c=a^2+b^2-k^2 \\
\text { i.e. centre of circle }(-g,-f)=[(a \cos \alpha+b \sin \alpha) \text {, } \\
(a \sin \alpha-b \cos \alpha)] \\
\text { and radius of circle }=\sqrt{g^2+f^2-c} \\
=\sqrt{(a \cos \alpha+b \sin \alpha)^2+(a \sin \alpha-b \cos \alpha)^2-a^2-b^2+k^2} \\
=\sqrt{\begin{array}{r}
a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha+2 a b \sin \alpha \cos \alpha+a^2 \sin ^2 \alpha \\
+b^2 \cos ^2 \alpha-2 a b \sin \alpha \cos \alpha-a^2-b^2+k^2
\end{array}} \\
=\sqrt{a^2\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+b^2\left(\sin ^2 \alpha+\cos ^2 \alpha\right)-a^2-b^2+k^2} \\
=\sqrt{a^2+b^2-a^2-b^2+k^2}=\sqrt{k^2}=k
\end{array}
\]

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Question 23 Marks

Find the equation of the circle passing through points $(a, a),(0, a)$ and $(a, 0)$.

Answer

Let the equation of circle be
$
(x-h)^2+(y-k)^2=r^2
$
As this circle passes through points $(a, a),(0, a)$ and $(a, 0)$. So,
$
\begin{array}{l}
(a-h)^2+(a-k)^2=r^2.......(1)\\
(0-h)^2+(a-k)^2=r^2.......(2) \\
(a-h)^2+(0-k)^2=r^2......(3)
\end{array}
$
On equating equation (1) and (2)
$
\begin{aligned}
(a-h)^2+(a-k)^2 & =(0-h)^2+(a-k)^2 \\
(a-h)^2 & =h^2 \\
a^2-2 a h+h^2 & =h^2 \quad \therefore h=\frac{1}{2} a
\end{aligned}
$
On equating equation (1) and (3)
$
\begin{array}{rlrl}
& (a-h)^2+(a-k)^2=(a-h)^2+k^2 \\
\Rightarrow & a^2-2 a k+k^2=k^2 & \therefore k=\frac{1}{2} a
\end{array}
$
Putting values of $h$ and $k$ in equation (1)
$
\begin{array}{rlrl}
\left(a-\frac{1}{2} a\right)^2+\left(a-\frac{1}{2} a\right)^2 & =r^2 \\
\Rightarrow \quad & \frac{1}{4} a^2+\frac{1}{4} a^2 & =r^2 \quad \therefore \quad r=\frac{1}{\sqrt{2}} a
\end{array}
$
So, equation of circle
$
\begin{aligned}
& \left(x-\frac{1}{2} a\right)^2+\left(y-\frac{1}{2} a\right)^2=\left(\frac{1}{\sqrt{2}} a\right)^2 \\
\Rightarrow \quad & x^2+y^2=a(x+y) \text { }
\end{aligned}
$

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Question 33 Marks

Write the coordinate centre of the ellipse $\frac{x^2-a x}{a^2}+\frac{y^2-b y}{b^2}=0$

Answer

$
\frac{x^2-a x}{a^2}+\frac{y^2-b y}{b^2}=0
$
Using completing square method :
$
\begin{array}{ll}
\Rightarrow & \frac{x^2-a x+\frac{a^2}{4}-\frac{a^2}{4}}{a^2}+\frac{y^2-b y+\frac{b^2}{4}-\frac{b^2}{4}}{b^2} \\
\Rightarrow & \frac{\left(x-\frac{a}{2}\right)^2-\frac{a^2}{4}}{a^2}+\frac{\left(y-\frac{b}{2}\right)^2-\frac{b^2}{4}}{b^2} \\
\Rightarrow & \frac{\left(x-\frac{a}{2}\right)^2}{a^2}-\frac{1}{4}+\frac{\left(y-\frac{b}{2}\right)^2}{b^2}-\frac{1}{4}=0 \\
\Rightarrow & \frac{\left(x-\frac{a}{2}\right)^2}{a^2}+\frac{\left(y-\frac{b}{2}\right)^2}{b^2}=\frac{1}{2} \\
\Rightarrow & \frac{\left(x-\frac{a}{2}\right)^2}{\left(\frac{a}{\sqrt{2}}\right)^2}+\frac{\left(y-\frac{b}{2}\right)^2}{\left(\frac{b}{\sqrt{2}}\right)^2}=1
\end{array}
$
Let $x-\frac{a}{2}= X , y-\frac{b}{2}= Y$
So,
\[
\frac{X^2}{\left(\frac{a}{\sqrt{2}}\right)^2}+\frac{Y^2}{\left(\frac{b}{\sqrt{2}}\right)^2}=1
\]
So coordinates of the centre $(0,0)$
$
\Rightarrow \quad X=0, Y=0, \quad \text { So, } x=\frac{1}{2} a, y=\frac{1}{2} b
$
So coordinates of centre of the ellipse $\left(\frac{a}{2}, \frac{b}{2}\right)$

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Question 43 Marks

If $e$ and $e^{\prime}$ be the eccentricity of a hyperbola and its conjugate, then prove that $\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1$.

Answer

Let the equation of the hyperbola be :
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\quad\quad\ldots\ldots (1)$
Equation of its conjugate hyperbola
$-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\quad\quad\ldots\ldots (2)$
Eccentricity of equation (1)
$\begin{array}{l}e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2} \\e^2=\frac{a^2+b^2}{a^2}\quad\quad\ldots\ldots\ (3)\end{array}$
Eccentricity of equation (2)
$\begin{array}{l}\left(e^{\prime}\right)^2=1+\frac{a^2}{b^2} \\\left(e^{\prime}\right)^2=\frac{b^2+a^2}{b^2}\quad\quad\ldots\ldots (4)\end{array}$
From equations (3) and (4)
$\begin{array}{l}\quad\quad\frac{1}{e^2}+\frac{1}{e^{\prime 2}} = \frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1 \\\Rightarrow \quad \frac{1}{e^2}+\frac{1}{e^{\prime 2}}=1 \end{array}$

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Question 53 Marks

Find the eccentricity of an ellipse having centre at the origin, axes along the coordinate axes and whose minor axis is equal to the distance between foci.

Answer

Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
So foci $F _1(a e, 0)$ and $F _2(-a e, 0)$
Minor axis $=2 b$
Distance between foci $F _1 F_2=2 a e$
According to question,
$2 a e=2 b$
$\Rightarrow \quad e=\frac{b}{a}\ldots\ldots (1)$
$\because \quad b^2=a^2\left(1-e^2\right)$
$\Rightarrow \quad \frac{b^2}{a^2}=1-e^2$
$\left(\frac{b}{a}\right)^2=1-e^2$
From equation (1),
$(e)^2=1-e^2$
$\begin{aligned} \Rightarrow \quad 2 e^2 & =1 \\ e^2 & =\frac{1}{2}\end{aligned}$
$\Rightarrow \quad e=\frac{1}{\sqrt{2}}$

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Question 63 Marks

Find the equation of the ellipse with axes as the axes of coordinates having latus rectum 5 and eccentricity $e=\frac{2}{3}$.

Answer

Equation of ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ldots\ldots (1)$
Here length of latus rectum $=5$
$\therefore \quad \frac{2 b^2}{a}=5\ldots\ldots (2)$
$\begin{array}{lrl}\Rightarrow & 2 b^2 & =5 a \\ \text { But } & b^2 & =a^2\left(1-e^2\right)\end{array}$
$\begin{array}{lr}\Rightarrow & 2 a^2\left(1-e^2\right)=5 a \\ \Rightarrow & 2 a\left(1-\frac{4}{9}\right)=5 \\ \Rightarrow & 2 a \times \frac{5}{9}=5\end{array}$
$\therefore \quad a=\frac{9}{2}$
Putting value of $a$ in equation (2)
$2 b^2=5\left(\frac{9}{2}\right)$
$\Rightarrow \quad b^2=\frac{45}{4}$
$\therefore \quad b=\frac{3 \sqrt{5}}{2}$
Now putting value of $a$ and $b$ in equation (1)
$\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$
$\Rightarrow \quad \frac{4 x^2}{81}+\frac{4 y^2}{45}=1$

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Question 73 Marks

Find the eccentricity of that ellipse, whose latus rectum is equal to the distance between foci.

Answer

Let the equation of ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
So, foci $F _1(a e, 0)$ and $F _2(-a e, 0)$
Distance between foci $F _1 F_2=2 a e$
Length of latus rectum $=\frac{2 b^2}{a}$
According to question, $\frac{2 b^2}{a}=2 a e$
$\begin{array}{ll} & b^2=a^2 e\ldots\ldots (1) \\ \text { but } & b^2=a^2\left(1-e^2\right)\ldots\ldots (2)\end{array}$
Solving equations (1) and (2)
$a^2 e=a^2\left(1-e^2\right)$
$\begin{array}{rlrl}\Rightarrow & e^2+e-1 =0 \\ \Rightarrow & e=\frac{-1 \pm \sqrt{1+4}}{2} \\ \Rightarrow & e=\frac{-1+\sqrt{5}}{2}\end{array}$
$e=\frac{-1-\sqrt{5}}{2} \quad$ (rejected)

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Question 83 Marks

Find the equation of that hyperbola whose latus rectum is 8 and conjugate axis is half of distance between foci.

Answer

According to question,
latus rectum $8=\frac{2 b^2}{a}$
$b^2=4 a\quad\ldots\ldots (1)$
and conjugate axis $2 b=\frac{1}{2}(2 a e)=a e$.
$2b=ae\quad\ldots\ldots (2)$
We know that $\quad b^2=a^2\left(e^2-1\right)$
From equation (1), $4 a=a^2\left(e^2-1\right)$
$\Rightarrow e^2=\frac{4}{a}+1\quad\ldots\ldots (3)$
On squaring equation (2) and then putting value of $b^2$ in equation (1),
$4 \times(4 a)=a^2 e^2$
$\Rightarrow \quad e^2=\frac{16}{a}\quad\ldots\ldots (4)$
From equations (3) and (4)
$ \frac{4}{a}+1=\frac{16}{a} $
$\frac{12}{a}=1 \Rightarrow a=12$
From equation (1), $b^2=4 \times 12 \Rightarrow b^2=48$
Now equation of hyperbola
$\frac{x^2}{(12)^2}-\frac{y^2}{48}=1$
$\begin{array}{ll}\Rightarrow & \frac{x^2}{144}-\frac{y^2}{48}=1 \\ \Rightarrow & x^2-3 y^2=144\end{array}$

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Question 93 Marks

Find the equation of that ellipse whose coordinates of vertices and foci are $( \pm 5,0)$ and $( \pm 4,0)$ respectively.

Answer

Equation of ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
The coordinates of vertices and foci are ( $\pm a, 0$ ) and $( \pm a e, 0)$ respectively,
$\begin{aligned}\text {So,}\quad\pm a & = \pm 5 \\\therefore a & =5 \\\text{and} \pm a e & = \pm 4 \\\therefore a e & =4\end{aligned}$
$\begin{aligned}\text {but}\quad a & =5 \\5 \times e & =4 \\e & =\frac{4}{5}\end{aligned}$
$\begin{aligned}\text {Again from}\quad e & =\sqrt{1-\frac{b^2}{a^2}} \\\frac{4}{5} & =\sqrt{1-\frac{b^2}{(5)^2}} \\\frac{16}{25} & =1-\frac{b^2}{25} \Rightarrow \frac{b^2}{25}=1-\frac{16}{25} \\\Rightarrow \frac{b^2}{25} & =\frac{9}{25} \Rightarrow b^2=9 \\b & =3\end{aligned}$
Putting values of $a$ and $b$ in equation (1),
$\frac{x^2}{25}+\frac{y^2}{9}=1$
This is the equation of required ellipse.

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