MATHS M.C.Q (1 Marks)

4. $\frac{4}{\sqrt{3}}$

Solution:

$3\text{x}^2+\text{y}^2=12$

$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{12}=1$

$\therefore\text{ a}^2=4$

$\Rightarrow\text{a}=2$

and $\text{b}^2=12$

$\Rightarrow\text{b}=2\sqrt{3}$

Since b > a, length of latus rectum $=\frac{2\text{a}^2}{\text{b}}=\frac{2\times4}{2\sqrt{3}}=\frac{4}{\sqrt{3}}$

3. x² + y² = 4a²

Solution:

Let ABC be an equilateral triangle in which mediam AD = 3a

Centre of the circle is same as the centroid of the triangle i.e., (0, 0)

AG : GD = 2 : 1

So, $\text{AG}=\frac{2}{3}\text{AD}=\frac{2}{3}\times3\text{a}=2\text{a}$

$\therefore$ The equation of the circle is,

(x - 0)² + (y - 0)² = (2a)²

⇒ x² + y² = 4a²

3. x² + y² - 6x - 6y + 9 = 0

Solution:

Given that the circle touches both axes.

Therefore, equation of the circle is, (x - a)² + (y - a)² = a²

Circle passes through the point (3, 6)

$\therefore$ (3 - a)² + (6 - a)² = a²

⇒ a² - 18a + 45 = 0

⇒ (a - 3)(a - 15) = 0

$\therefore$ a = 3, a = 15

For a = 3, the equation of circle is,

(x - 3)² + (y - 3)² = 9

⇒ x² + y² - 6x - 6y + 9 = 0

1. y² = 8(x + 3)

Solution:

Given that vertex $\equiv(-3,0)$ and directrix, x + 5 = 0

So, focus $\equiv\text{S}(-1,0)$

For any point of parabola P(x, y) we have,

$\text{SP}=\text{PM}$

$\Rightarrow\sqrt{(\text{x}+1)+\text{y}^2}=|\text{x}+5|$

$\Rightarrow\text{x}^2+2\text{x}+1+\text{y}^2=\text{x}^2+10\text{x}+25$

$\Rightarrow\text{y}^2=8\text{x}+24$

$\Rightarrow\text{y}^2=8(\text{x}+3)$

2.   a² = b²(1 - e²)

Solution:

Given equation is $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b})$

$\therefore\text{ Eccentricity e}=\sqrt{1-\frac{\text{a}^2}{\text{b}^2}}$

$\Rightarrow\text{e}^2=1-\frac{\text{a}^2}{\text{b}^2}$

$\Rightarrow\frac{\text{a}^2}{\text{b}^2}=(1-\text{e}^2)$

$\Rightarrow\text{a}^2=\text{b}^2(1-\text{e}^2)$  

1. $\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$

Solution:

Given that $\text{e}=\frac{3}{2}$

and foci $=(\pm\text{ae},0)=(\pm2,0)$

$\therefore\ \text{ae}=2$

$\text{a}\times\frac{3}{2}=2$

$\Rightarrow\text{a}=\frac{4}{3}$

Now we know that $\text{b}^2=\text{a}^2(\text{e}^2-1)$

$\text{b}^2=\frac{16}{9}\Big(\frac{9}{4}-1\Big)$

$\Rightarrow\text{b}^2=\frac{16}{9}\times\frac{5}{4}$

$\Rightarrow\text{b}^2=\frac{20}{9}$

So, the equation of the hyperbola is,

$\frac{\text{x}^2}{\big(\frac{4}{3}\big)^2}-\frac{\text{y}^2}{\frac{20}{9}}=1$

$\Rightarrow\frac{9\text{x}^2}{16}-\frac{9\text{y}^2}{20}=1$

$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{20}=\frac{1}{9}$

$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$

3. $\frac{2}{\sqrt{3}}$

Solution:

Let the equation of the hyperbola be $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$

Length of latus rectum = 8

$\therefore\ \frac{2\text{b}^2}{2}=8$

$\Rightarrow\text{b}^2=4\text{a}$

Conjugate axis = half of the distance between the foci

$\therefore\ 2\text{b}=\text{ae}$

Now, $\text{b}^2=\text{a}^2(\text{e}^2-1)$

From eqs. (i) and (iii), we get

$\frac{\text{a}^2\text{e}^2}{4}=\text{a}^2(\text{e}^2-1)$

$\Rightarrow\text{e}^2=4\text{e}^2-4$

$\Rightarrow\text{e}^2=\frac{4}{3}$

$\Rightarrow\text{e}=\frac{2}{\sqrt{3}}$

1. 7x² + 2xy + 7y² - 10x + 10y + 7 = 0

Solution:

Given that, fouus of the ellipse is S(1, -1) and the equation of directrix is x - y - 3 = 0

Also, $\text{e}=\frac{1}{2}$

From definition of ellipse, for any point P(x, y) on the ellipse, we have SP = ePM, where M is foot of the perpendicular from point P to the directrix.

$\therefore\ \sqrt{(\text{x}-1)^2+(\text{y}+1)^2}=\frac{1}{2}\frac{|\text{x}-\text{y}-3|}{\sqrt{2}}$

⇒ 8x² - 16x + 16 + 8y² + 16y = x² + y² + 9 - 2xy + 6y - 6x

⇒ 7x² + 2xy + 7y² - 10x + 10y + 7 = 0

3. $25\pi$

Solution:

Given that the centre of the circle is (1, 2)

Radius of the circle $=\sqrt{(4-1)^2+(6-2)^2}$

$=\sqrt{9+16}=5$

So, the area of the circle $=\pi\text{r}^2$

$=\pi\times(5)^2=25\pi$

1. $\text{x}^2-\text{y}^2=32$

Solution:

We know that distance between the foci = 2ae

$\therefore\ 2\text{ae}=16$

$\Rightarrow​​\text{ae}=8$

Given that $\text{e}=\sqrt{2}$

$\therefore\ \sqrt{2}\text{a}=8$

$\Rightarrow\text{a}=4\sqrt{2}$

Now, $\text{b}^2=\text{a}^2(\text{e}^2-1)$

$\Rightarrow\text{b}^2=32(32-1)$

$\Rightarrow\text{b}^2=32$

So, the equation of the hyperbola is,

$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$

$\Rightarrow\frac{\text{x}^2}{32}-\frac{\text{y}^2}{32}=1$

 $\Rightarrow\text{x}^2-\text{y}^2=32$

1.  x² = -12y

Solution:

According to the definition of parabola,

$\sqrt{(\text{x}-0)^2+(\text{y}+3)^2}=\Bigg|\frac{\text{y}-3}{\sqrt{(0)^2+(1)^2}}\Bigg|$

$\Rightarrow\sqrt{\text{x}^2+\text{y}^2+9+6\text{y}}=|\text{y}-3|$

Squaring both sides, we get

x² + y² + 9 + 6y = y² + 9 - 6y

⇒ x² + 9 + 6y = 9 - 6y

⇒ x² = -12y

2. $\frac{4}{3}$

Solution:

Given parabola is y² = 4ax

If the parabola is passing through (3, 2)

Then (2)² = 4a × 3

⇒ 4 = 12a

$\Rightarrow\text{a}=\frac{1}{3}$

Nowm length of the latus rectum $=4\text{a}=4\times\frac{1}{3}=\frac{4}{3}$

1. x² + y² + 13y = 0

Solution:

Let the equation of the circle be,

(x - h)² + (y - k)² = r²

Let the centre be (0, a)

$\therefore$ Radius r = a

So, the equation of the circle is

(x - 0)² + (y - a)² = a²

⇒ x² + (y - a)² = a²

⇒ x² + y² + a² - 2ay = a²

⇒ x² + y² - 2ay = 0 .....(i)

(image)

Now, CP = r

$\Rightarrow\sqrt{(2-0)^2+(3-\text{a}^2)}=\text{a}$

$\Rightarrow\sqrt{4+9+\text{a}^2-6\text{a}}=\text{a}$

$\Rightarrow\sqrt{13+\text{a}^2-6\text{a}}=\text{a}$

$\Rightarrow13+\text{a}^2-6\text{a}=\text{a}^2$

$\Rightarrow13-6\text{a}=0$

$\therefore\ \text{a}=\frac{13}{6}$

Putting the value of a in eq. (i) we get

$\text{x}^2+\text{y}^2-2\Big(\frac{13}{6}\Big)\text{y}=0$

$\Rightarrow3\text{x}^2+3\text{y}^2-3\text{y}=0$

Note: (a) option is correct and is should be (dout solution)