(Each question 2 marks)

Question 512 Marks

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Answer

The equation of circle is
$(x - h)^2 + (y - k)^2 = r^2 . . . (i)$
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2).
$\therefore$ radius of circle $= \sqrt {{{(4 - 2)}^2} + {{(5 - 2)}^2}} = \sqrt {4 + 9} = \sqrt {13}$
Now the equation of required circle is
$(x - 2)^2 + (y - 2)^2= (\sqrt{13})^2 \Rightarrow x^2 + 4 - 4x + y^2 + 4 - 4y = 13$
$\Rightarrow x^2 + y^2 - 4x - 4y - 5 = 0$

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Question 522 Marks

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Answer

The circle makes intercepts a with x -axis and b with y -axis.
$\therefore \mathrm{OA}=\mathrm{a} \text { and } \mathrm{OB}=\mathrm{b}$
So the co-ordinates of $A$ are $(a, 0)$ and $B$ are $(0, b)$
Now the circle passes through three points $O(0,0), A(a, 0)$ and $B(0, b)$
Putting the co-ordinates of three points in the equation of circle.
$\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \ldots$(i)
$\mathrm{c}=0$
$\mathrm{a}^2+2 \mathrm{ga}=0 \Rightarrow \mathrm{a}(\mathrm{a}+2 \mathrm{~g})=0 \Rightarrow g=\frac{-1}{2} a$
$\mathrm{~b}^2+2 \mathrm{fb}=0 \Rightarrow \mathrm{~b}(\mathrm{~b}+2 \mathrm{f})=0 \Rightarrow f=\frac{-1}{2} b$
Putting these values of $\mathrm{g}, \mathrm{f}$ and c in (i) we have
$x^2+y^2+2 \times \frac{-1}{2} a x+2 \times \frac{-1}{2} b y+0=0$
$\Rightarrow x^2+y^2-a x-b y=0$
which is required equation of circle.

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Question 532 Marks

Find the equation of the circle with centre (0, 2) and radius 2

Answer

Here $h=0, k=2$ and $r=2$
The equation of circle is
$(x-h)^2+(y-k)^2=r^2$
$\therefore(x-0)^2+(y-2)^2=(2)^2$
$\Rightarrow x^2+y^2+4-4 y=4$
$x^2+y^2-4 y=0$
Which is required equation of circle.

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Question 542 Marks

Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2, -3).

Answer

Since the parabola is symmetric about the y-axis and has its vertex at the origin, the equation is of the form $x^2 = 4ay$ or $x^2 = -4ay$,
But the parabola passes through (2,–3) which lies in the fourth quadrant, it must open downwards.
Thus the equation is of the form $x^2 = -4ay$
Since the parabola passes through ( 2, -3), we have
$2^{2}=-4 a(-3), \text { i.e., } a=\frac{1}{3}$
Therefore, the equation of the parabola is
$x^{2}=-4\left(\frac{1}{3}\right) y, $ i.e., $3 x^2 = -4y$

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Question 552 Marks

Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).

Answer

Given, the vertex is $(0,0)$ and focus is at $(0,2)$ which lies on $Y$-axis.
The $Y$ axis is the axis of parabola.
Therefore, equation of parabola is of the form
$x^2=4 a y$
$x^2=4(2) y \text { i.e., } x^2=8 y$

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Question 562 Marks

Find the equation of the parabola with focus (2, 0) and directrix x = – 2.

Answer

Given that the directrix is $x=-2$ and the focus is $(2,0)$,
Since the focus $(2,0)$ lies on the $x$-axis, the $x$-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form either $y^2=4 a x$ or $y^2=-4 a x$.
Since the directrix is $x=-2$ and the focus is $(2,0)$, the parabola is to be of the form $y^2=4 a x$ with $a=2$.
Hence the required equation is $y^2=4(2) x=8 x$

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Question 572 Marks

Find the coordinates of focus, axis, the equation of directrix and latus rectum of parabola $\mathrm{y}^2=8 \mathrm{x}$.

Answer

We have, equation of parabola is $y^2=8 x$.
The given equation involves $y^2$, so the axis of symmetry is along X -axis. The coefficient of x is positive, so the parabola opens to right.
On comparing with the given equation $y^2=4 a x$, we get,
$a=2$
Thus, focus $=(2,0)$
Equation of directrix, $x=-2$
Length of latus rectum is $4 a=4 \times 2=8$.

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Question 582 Marks

Find the equation of the circle which passes through the points (2, - 2) and (3, 4) and whose centre lies on the line x + y = 2.

Answer

Let the equation of circle with centre $(h, k)$ and radius $r$ be $(x-h)^2+(y-k)^2=r^2 \ldots(i)$
Since, circle passes through the points $(2,-2)$ and $(3,4)$, so the points $(2,-2)$ and $(3,4)$ will lie on Eq. (i).
$\therefore(2-h)^2+(-2-k)^2=r^2 \ldots \text { (iii) }$
$\text { and }(3-h)^2+(4-k)^2=r^2 \ldots \text { (iii) }$
Now, from Eqs. (ii) and (iii), we get
$(2-h)^2+(-2-k)^2=(3-h)^2+(4-k)^2$
$\Rightarrow 4+h^2-4 h+4+k^2+4 k=9+h^2-6 h+16+k^2-8 k$
$\Rightarrow 2 h+12 k=17 \ldots \text { (iv) }$
Also, given that centre ( $h, k$ ) lies on $x+y=2$. So, it will satisfy it.
$\therefore h+k=2 \ldots(v)$
On solving Eqs. (iv) and (v), we get
$\mathrm{h}=0.7, \mathrm{k}=1.3$
Now, $r^2=(2-0.7)^2+(-2-1.3)^2=1.69+10.89=12.58$
On putting $h=0.7, k=13$ and $r^2=12.58$ in Eq. (i), we get
$(x-0.7)^2+(y-1.3)^2=12.58$
which is the required equation of circle.

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Question 592 Marks

Find the centre and the radius of the circle: $x^2+y^2+8 x+10 y-8=0$

Answer

The given equation is $\left(x^2+8 x\right)+\left(y^2+10 y\right)=8$
Now, completing the squares within the parenthesis, we get
$\left(x^2+8 x+16\right)+\left(y^2+10 y+25\right)=8+16+25$
i.e. $(x+4)^2+(y+5)^2=49$
i.e. $\{\mathrm{x}-(-4)\}^2+\{\mathrm{y}-(-5)\}^2=49$ [comparing with $\{\mathrm{x}-(\mathrm{h})\}^2+\{\mathrm{y}-(\mathrm{k})\}^2=\mathrm{r}^2$, centre $(-\mathrm{h},-\mathrm{k})$ and r radius]
Therefore, the given circle has centre at $(-4,-5)$ and radius 7 .

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Question 602 Marks

A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse.

Answer

Let AB be the rod making an angle θ with OX as shown in figure and P (x, y) the point on it such that AP = 6 cm
Since AB = 15 cm, we have

PB = 9 cm.
From P draw PR and PQ perpendiculars on x-axis and y-axis, respectively.
From $\Delta \mathrm{PBQ}, \cos \theta=\frac{x}{9}$
From $\Delta \mathrm{PRA}, \sin \theta=\frac{y}{6}$
Since $\cos ^2 \theta+\sin ^2 \theta=1$
$\left(\frac{x}{9}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1$
or $\frac{x^{2}}{81}+\frac{y^{2}}{36}=1$
Therefore, locus of P is an ellipse.

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Question 612 Marks

Find the equation of the hyperbola with foci (0, $\pm$ 3) and vertices $\left(0, \pm \frac{\sqrt{11}}{2}\right)$

Answer

Since the foci is on the y-axis, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$given, foci (0, $\pm$ 3) and vertices $\left(0, \pm \frac{\sqrt{11}}{2}\right)$
Since vertices are $\left(0, \pm \frac{\sqrt{11}}{2}\right), \quad a=\frac{\sqrt{11}}{2}$
Also, since foci are (0, ± 3); c= ae = 3 and $b^2 = c^2 – a^2$ = $\frac{25}{4}$
Therefore, the equation of the hyperbola is $\frac{y^{2}}{\left(\frac{11}{4}\right)}-\frac{x^{2}}{\left(\frac{25}{4}\right)}$ = 1, i.e., $100 y^2 – 44 x^2 = 275$

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MATHS (Each question 2 marks)