2 Marks Questions

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 = r² . . . (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 = $(\sqrt{13})^2$ $\Rightarrow$ x² + 4 - 4x + y² + 4 - 4y = 13
$\Rightarrow$ x² + y² - 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.
$\therefore$ OA = a and OB = 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.
x² + y² + 2gx + 2fy + c = 0. . . (i)
c = 0
a² + 2ga = 0 $\Rightarrow$ a(a + 2 g) = 0 $\Rightarrow g = \frac{{ - 1}}{2}a$
b² + 2fb = 0 $\Rightarrow$ b (b + 2 f ) = 0 $\Rightarrow f = \frac{{ - 1}}{2}b$
Putting these values of g, f and c in (i) we have
${x^2} + {y^2} + 2 \times \frac{{ - 1}}{2}ax + 2 \times \frac{{ - 1}}{2}by + 0 = 0$
$\Rightarrow$ x² + y² - ax - by = 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 = r²
$\therefore$ (x - 0)² + (y - 2)² = (2)²
$\Rightarrow$ x² + y² + 4 - 4y = 4
x² + y² - 4y = 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 x² = 4ay or x² = -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 x² = -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 x² = -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
x² = 4ay
x² = 4(2)y i.e., x² = 8y

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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 y² = 4ax or y² = -4ax.
Since the directrix is x = – 2 and the focus is (2, 0),
the parabola is to be of the form y² = 4ax with a = 2.
Hence the required equation is y²= 4(2)x = 8x

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Question 572 Marks

Find the coordinates of focus, axis, the equation of directrix and latus rectum of parabola y² = 8x.

Answer

We have, equation of parabola is y² = 8x.
The given equation involves y², 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 y² = 4ax, we get,
a = 2
Thus, focus = (2, 0)
Equation of directrix, x = - 2
Length of latus rectum is 4a = 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 = r² ...(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 = r² ...(ii)
and (3 - h)² + (4 - k)² = r²...(iii)
Now, from Eqs. (ii) and (iii), we get
(2 - h)² + (- 2 - k)² = (3 - h)² + (4 - k)²
$\Rightarrow$ 4 + h² - 4h + 4 + k² + 4k = 9 + h² - 6h + 16 + k²- 8k
$\Rightarrow$ 2h + 12k = 17 ...(iv)
Also, given that centre (h, k) lies on x + y = 2. So, it will satisfy it.
$\therefore$ h + k = 2 ...(v)
On solving Eqs. (iv) and (v), we get
h = 0.7, k = 1.3
Now, r² = (2 - 0.7)² + (- 2 - 1.3)² = 1.69 + 10.89 = 12.58
On putting h = 0.7, k = 13 and r² = 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: x² + y² + 8x + 10y – 8 = 0

Answer

The given equation is (x² + 8x) + (y² + 10y) = 8
Now, completing the squares within the parenthesis, we get
(x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25
i.e. (x + 4)² + (y + 5)² = 49
i.e. {x – (– 4)}² + {y – (–5)}² = 49 [comparing with {x – (h)}² + {y – (k)}² = r², centre(-h, -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 cos² θ + sin² θ = 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 b² = c² – a² = $\frac{25}{4}$
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 y² – 44 x² = 275

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MATHS 2 Marks Questions