M.C.Q (1 Marks)

MCQ 11 Mark

The eccentricity of the hyperbola whose latus rectum is $8$ and conjugate axis is equal to half of the distance between the foci is:

A
$\frac{4}{3}$
B
$\frac{4}{\sqrt{3}}$
✓
$\frac{2}{\sqrt{3}}$
D
none of these.

Answer

Correct option: C.

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

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
$\frac{\text{a}^2\text{e}^2}{4}=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\text{e}^2=4\text{e}^2-4$
$\Rightarrow\text{e}^2=\frac{4}{3}$
$\Rightarrow\text{e}=\frac{2}{\sqrt{3}}$

View full question & answer→

MCQ 21 Mark

The equation of the ellipse whose focus is $(1, -1),$ the directrix the line $x - y - 3 = 0$ and eccentricity $\frac{1}{2}$ is:

✓
$7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0$
B
$7x^2 + 2xy + 7y^2 + 7 = 0$
C
$7x^2 + 2xy + 7y^2 + 10x - 10y - 7 = 0$
D
none

Answer

Correct option: A.

$7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0$

$7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0$
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.
$\therefore\ \sqrt{(\text{x}-1)^2+(\text{y}+1)^2}=\frac{1}{2}\frac{|\text{x}-\text{y}-3|}{\sqrt{2}}$
$\Rightarrow 8x^2- 16x + 16 + 8y^2 + 16y = x^2 + y^2 + 9 - 2xy + 6y - 6x$
$\Rightarrow 7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0$

View full question & answer→

MCQ 31 Mark

The area of the circle centred at $(1, 2)$ and passing through $(4, 6)$ is:

A
$5\pi$
B
$10\pi$
✓
$25\pi$
D
none of these.

Answer

Correct option: C.

$25\pi$

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 $=\pi\text{r}^2$
$=\pi\times(5)^2$
$=25\pi$

View full question & answer→

MCQ 41 Mark

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length $3a$ is:
$[$Hint: Centroid of the triangle coincides with the centre of the circle and the radius of the circle is $\frac{2}{3}$ of the length of the mediam$]$

A
$x^2 + y^2 = 9a^2$
B
$x^2 + y^2 = 16a^2$
✓
$x^2 + y^2 = 4a^2$
D
$x^2 + y^2= a^2$

Answer

Correct option: C.

$x^2 + y^2 = 4a^2$

$x^2 + y^2 = 4a^2$
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$,$
$(x - 0)^2 + (y - 0)^2= (2a)^2$
$\Rightarrow x^2 + y^2 = 4a^2$

View full question & answer→

MCQ 51 Mark

Equation of a circle which passes through $(3, 6)$ and touches the axes is:

A
$x^2 + y^2 + 6x + 6y + 3 = 0$
B
$x^2 + y^2 - 6x - 6y - 9 = 0$
✓
$x^2 + y^2 - 6x - 6y + 9 = 0$
D
none of these.

Answer

Correct option: C.

$x^2 + y^2 - 6x - 6y + 9 = 0$

$x^2 + y^2 - 6x - 6y + 9 = 0$
Given that the circle touches both axes.
Therefore$,$ equation of the circle is$, (x - a)^2 + (y - a)^2 = a^2$
Circle passes through the point $(3, 6)$
$\therefore (3 - a)^2 + (6 - a)^2 = a^2$
$\Rightarrow a^2 - 18a + 45 = 0$
$\Rightarrow (a - 3)(a - 15) = 0$
$\therefore\ = 3, a = 15$
For $a = 3,$ the equation of circle is,
$(x - 3)^2 + (y - 3)^2 = 9$
$\Rightarrow x^2 + y^2 - 6x - 6y + 9 = 0$

View full question & answer→

MCQ 61 Mark

If the focus of a parabola is $(0, -3)$ and its directrix is $y = 3,$ then its equation is:

✓
$x^2 = -12y$
B
$x^2= 12y$
C
$y^2 = -12x$
D
$y^2= 12x$

Answer

Correct option: A.

$x^2 = -12y$

$x^2 = -12y$
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
$x^2 + y^2 + 9 + 6y = y^2 + 9 - 6y$
$\Rightarrow x^2 + 9 + 6y = 9 - 6y$
$\Rightarrow x^2 = -12y$

View full question & answer→

MCQ 71 Mark

If the vertex of the parabola is the point $(-3, 0)$ and the directrix is the line $x + 5 = 0,$ then its equation is:

✓
$y^2 = 8(x + 3)$
B
$x^2 = 8(y + 3)$
C
$y^2= -8(x + 3)$
D
$y^2 = 8(x + 5)$

Answer

Correct option: A.

$y^2 = 8(x + 3)$

$y^2 = 8(x + 3)$
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,
$\text{SP}=\text{PM}$
$\Rightarrow\sqrt{(\text{x}+1)+\text{y}^2}=|\text{x}+5|$
$\Rightarrow\text{x}^2+2\text{x}+1+\text{y}^2=\text{x}^2+10\text{x}+25$
$\Rightarrow\text{y}^2=8\text{x}+24$
$\Rightarrow\text{y}^2=8(\text{x}+3)$

View full question & answer→

MCQ 81 Mark

If the parabola $y^2 = 4ax$ passes through the point $(3, 2)$, then the length of its latus rectum is:

A
$\frac{2}{3}$
✓
$\frac{4}{3}$
C
$\frac{1}{3}$
D
$4$

Answer

Correct option: B.

$\frac{4}{3}$

Given parabola is $y^2 = 4ax$
If the parabola is passing through $(3, 2)$
Then $(2)^2 = 4a \times 3$
$\Rightarrow 4 = 12a$
$\Rightarrow\text{a}=\frac{1}{3}$
Nowm length of the latus rectum $=4\text{a}=4\times\frac{1}{3}=\frac{4}{3}$

View full question & answer→

MCQ 91 Mark

The distance between the foci of a hyperbola is $16$ and its eccentricity is $2.$ Its equation is:

✓
$\text{x}^2-\text{y}^2=32$
B
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
C
$2\text{x}-3\text{y}^2=7$
D
none of these.

Answer

Correct option: A.

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

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

View full question & answer→

MCQ 101 Mark

If e is the eccentricity of the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b}),$ then:

A
$b^2= a^2(1 - e^2)$
B
$a^2= b^2(1 - e^2)$
✓
$a^2= b^2(e^2 - 1)$
D
$b^2= a^2(e^2 - 1)$

Answer

Correct option: C.

$a^2= b^2(e^2 - 1)$

Given equation is $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b})$
$\therefore\text{Eccentricity e}=\sqrt{1-\frac{\text{a}^2}{\text{b}^2}}$
$\Rightarrow\text{e}^2=1-\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow\frac{\text{a}^2}{\text{b}^2}=(1-\text{e}^2)$
$\Rightarrow\text{a}^2=\text{b}^2(1-\text{e}^2)$

View full question & answer→

MCQ 111 Mark

Equation of the hyperbola with eccentricty $\frac{3}{2}$ and foci at $(\pm2,0)$ is:

✓
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$
B
$\frac{\text{x}^2}{9}-\frac{\text{y}^2}{9}=\frac{4}{9}$
C
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
D
none of these.

Answer

Correct option: A.

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

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,
$\frac{\text{x}^2}{\big(\frac{4}{3}\big)^2}-\frac{\text{y}^2}{\frac{20}{9}}=1$
$\Rightarrow\frac{9\text{x}^2}{16}-\frac{9\text{y}^2}{20}=1$
$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{20}=\frac{1}{9}$
$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$

View full question & answer→

MCQ 121 Mark

Equation of the circle with centre on the $y-$axis and passing through the origin and the point $(2, 3)$ is:

✓
$x^2 + y^2 + 13y = 0$
B
$3x^2 + 3y^2 + 13x + 3 = 0$
C
$6x^2+ 6y^2 - 13x = 0$
D
$x^2 + y^2 + 13x + 3 = 0$

Answer

Correct option: A.

$x^2 + y^2 + 13y = 0$

Let the equation of the circle be,
$(x - h)^2 + (y - k)^2 = r^2$
Let the centre be $(0, a)$
$\therefore$ Radius $r = a$
So, the equation of the circle is
$(x - 0)^2 + (y - a)^2 = a^2$
$\Rightarrow x^2 + (y - a)^2 = a^2$
$\Rightarrow x^2 + y^2 + a^2 - 2ay = a^2$
$\Rightarrow x^2 + y^2 - 2ay = 0 .....(i)$
$($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$)$

View full question & answer→

MCQ 131 Mark

The length of the latus rectum of the ellipse $3x^2 + y^2 = 12$ is:

A
$4$
B
$3$
C
$8$
✓
$\frac{4}{\sqrt{3}}$

Answer

Correct option: D.

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

$\frac{4}{\sqrt{3}}$
$3\text{x}^2+\text{y}^2=12$
$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{12}=1$
$\therefore\text{ a}^2=4$
$\Rightarrow\text{a}=2$
and $\text{b}^2=12$
$\Rightarrow\text{b}=2\sqrt{3}$
Since $b > a$, length of latus rectum $=\frac{2\text{a}^2}{\text{b}}=\frac{2\times4}{2\sqrt{3}}=\frac{4}{\sqrt{3}}$

View full question & answer→

Maths M.C.Q (1 Marks)

Test yourself on this topic