MCQ

MCQ 2012 Marks

If the origin is shifted to the point (-3, 2), axes remaining parallel, the new co-ordinate of the point is (3, 1), then its old co-ordinate will be

A
(0,1)
B
(0,4)
C
(1,0)
✓
(0,3)

Answer

Correct option: D.

(0,3)

(D)
$\begin{array}{l}(h, k) \equiv(-3,2),(X, Y) \equiv(3,1) \\∴ x=X+h \text { and } y=Y+k \\∴ x=3-3=0, y=1+2=3\end{array}$
∴ Old co-ordinate $\equiv(0,3)$

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MCQ 2022 Marks

The new co-ordinates of a point (4, 5), when the origin is shifted to the point (1, -2), are

A
(5, 3)
B
(3, 5)
✓
(3, 7)
D
None of these

Answer

Correct option: C.

(3, 7)

(C)
We know that, if the origin is shifted to $(h, k)$, then new co-ordinates of a point $(x, y)$ becomes $(x- h , y- k )$.
Therefore, the new co-ordinates of $(4,5)$ with respect to new origin $(1, -2)$ are $(3, 7)$.

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MCQ 2032 Marks

Length of intercept on X axis of the locus $x^2+y^2-4 x-6 y-12=0$ is

A
5 units
B
2 units
✓
8 units
D
15 units

Answer

Correct option: C.

8 units

(C)
On $X -$ axis, $y=0$
Putting $y=0$ in the equation
$x^2+y^2-4 x-6 y-12=0$, we get
$x=6$ or $x=-2$
∴ Point on X -axis will be $(-2,0)$ and $(6,0)$
∴ Length of intercept $=\sqrt{(6+2)^2}=\sqrt{64}=8$ units

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MCQ 2042 Marks

If O is the origin and Q is a variable point on $x^2=4 y$, then locus of the midpoint of seg OQ is

✓
$x^2=2 y$
B
$x^2=4 y$
C
$y^2=2 x$
D
$y^2=4 x$

Answer

Correct option: A.

$x^2=2 y$

(A)
Let $P \equiv(x, y)$ be the midpoint of seg OQ and
$Q \equiv( h , k )$
Since, Q lies on the locus $x^2=4 y$
$h^2=4 k$... (i)
But, $P \equiv\left(\frac{ h }{2}, \frac{ k }{2}\right)$
$\begin{array}{l}x=\frac{h}{2}, y=\frac{k}{2} \\\Rightarrow h=2 x, k=2 y\end{array}$
Putting h and k in (i), we get
$(2 x)^2=4(2 y) \Rightarrow 4 x^2=8 y$
Equation of locus is $x^2=2 y$

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MCQ 2052 Marks

Which of the following points lie on the locus $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ?$

A
$\left(a \sin ^2 \theta, b \cos ^2 \theta\right)$
✓
$( a \cos \theta, b \sin \theta)$
C
$\left(\frac{a}{\sin \theta}, \frac{b}{\cos \theta}\right)$
D
$\left( a ^2 \sin \theta, b^2 \cos \theta\right)$

Answer

Correct option: B.

$( a \cos \theta, b \sin \theta)$

(B)
By substituting each of the points in the equation $\frac{x^2}{ a ^2}+\frac{y^2}{b^2}=1$, we get only point $(a \cos \theta, b \sin \theta)$ satisfies it.

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MCQ 2062 Marks

The equation of a locus of the point P (x, y) if $x= a \cos \theta+ b \sin \theta, y= a \sin \theta- b \cos \theta$ is

✓
$x^2+y^2=a^2+b^2$
B
$x^2+y^2= a ^2$
C
$x^2+y^2=2 a ^2$
D
$x^2+y^2=a^2-b^2$

Answer

Correct option: A.

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

(A)
$x=a \cos \theta+b \sin \theta$...(i)
$y=a \sin \theta-b \cos \theta$...(ii)
Squaring and adding (i) and (ii), we get
$x^2+y^2= a ^2+ b ^2$

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MCQ 2072 Marks

If points (-4, 4) and (-16, b) lie on the locus $y^2=a x$, then

A
$a= \pm 8, b= \pm 4$
✓
$a=-4, b= \pm 8$
C
$a=4, b=-4$
D
$a=-4, b=-16$

Answer

Correct option: B.

$a=-4, b= \pm 8$

(B)
Since, $(-4,4)$ lies on $y^2=a x$
$\begin{array}{l}∴(4)^2=a(-4)
\\\Rightarrow a=-4\end{array}$
Also, $(-16, b)$ lies on $y^2= a x$
∴ $b^2=-4(-16)=64$
$\Rightarrow b= \pm 8$

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MCQ 2082 Marks

If points (3, 2) and (-1, -2) lie on the locus ax + by = 5 then a and b are

A
a = 2, b = 3
B
a = 5, b = 2
C
a = - 5, b = 3
✓
a = 5, b = -5

Answer

Correct option: D.

a = 5, b = -5

(D)
Putting $x=3, y=2$ in equation $a x+ b y=5$,
we get
$3 a+2 b=5$...(i)
Putting $x=-1, y=-2$ in equation $a x+ b y=5$,
we get
$-a-2 b=5$...(ii)
Solving (i) and (ii), we get
$a =5, b=-5$

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MCQ 2092 Marks

For which value of K, point (-8, 6) lies on the locus $\frac{x^2}{4}+\frac{y^2}{3}= K ?$

A
K = 2
B
K = - 6
✓
K = 28
D
K = - 9

Answer

Correct option: C.

K = 28

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MCQ 2102 Marks

Which of the following points lies on the locus $2 x^2+2 y^2-5 x+11 y-3=0 ?$

A
(1,2)
B
(2, 1)
C
(7,5)
✓
(4,-1)

Answer

Correct option: D.

(4,-1)

(D)
Substituting each pair of the points in the equation $2 x^2+2 y^2-5 x+11 y-3=0$, we get only point $(4,-1)$ satisfies it.

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MCQ 2112 Marks

The locus of a point whose difference of distances from the points (3, 0) and (-3, 0) is 4 is

✓
$\frac{x^2}{4}-\frac{y^2}{5}=1$
B
$\frac{x^2}{5}-\frac{y^2}{4}=1$
C
$\frac{x^2}{2}-\frac{y^2}{3}=1$
D
$\frac{x^2}{3}-\frac{y^2}{2}=1$

Answer

Correct option: A.

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

(A)
Required equation of locus is
$\left|\sqrt{(x-3)^2+y^2}-\sqrt{(x+3)^2+y^2}\right|=4$
$\Rightarrow \sqrt{(x-3)^2+y^2}=\sqrt{(x+3)^2+y^2} \pm 4$
$\Rightarrow(x-3)^2+y^2=16+(x+3)^2+y^2$$\pm 8 \sqrt{(x+3)^2+y^2}$
$\Rightarrow-12 x-16= \pm 8 \sqrt{(x+3)^2+y^2}$
$\Rightarrow(3 x+4)^2=4(x+3)^2+4 y^2$
$\Rightarrow 5 x^2-4 y^2=20$
$\Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=1$

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MCQ 2122 Marks

The locus of a point which is equidistant from (1, 2) and the X- axis is

A
$x^2-2 x-6 y+10=0$
B
$x^2+2 x-6 y+10=0$
✓
$x^2-2 x-4 y+5=0$
D
$x^2+2 x-4 y-5=0$

Answer

Correct option: C.

$x^2-2 x-4 y+5=0$

(C)
Let $P (x, y)$ be any point on the locus and
$A \equiv(1,2)$
According to the given condition,
$\begin{array}{l}l(AP)=y \\
\sqrt{(x-1)^2+(y-2)^2}=y\end{array}$
Squaring both sides, we get
$\begin{array}{l}(x-1)^2+(y-2)^2=y^2 \\\Rightarrow x^2-2 x-y+5=0\end{array}$

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MCQ 2132 Marks

What is the equation of the locus of a point which moves such that 4 times its distance from the X-axis is the square of its distance from the origin?

A
$x^2-y^2-4 y=0$
✓
$x^2+y^2-4|y|=0$
C
$x^2+y^2-4 x=0$
D
$2^2+y^2-4|x|=0$

Answer

Correct option: B.

$x^2+y^2-4|y|=0$

(B)Let $\left(x_1, y_1\right)$ be any point on the locus.
According to the given condition,
$\begin{array}{l}4\left|y_1\right|=x_1^2+y_1^2 \\ \Rightarrow x_1^2+y_1^2-4\left|y_1\right|=0\end{array}$
∴ equation of the locus is
$x^2+y^2-4|y|=0$

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MCQ 2142 Marks

The equation of a locus of the point whose distance from Y-axis is two times its distance from the origin is

✓
$3 x^2+4 y^2=0$
B
$x^2+2 y^2=0$
C
$2 y^2+x^2=0$
D
$4 x^2+3 y^2=0$

Answer

Correct option: A.

$3 x^2+4 y^2=0$

(A)
Let $P (x, y)$ be any point on the locus.
According to the given condition,
distance of P from Y -axis $=2( OP )$
$\begin{array}{l}\Rightarrow x=2 \sqrt{(x-0)^2+(y-0)^2} \\
\Rightarrow x=2 \sqrt{x^2+y^2} \\\Rightarrow x^2=4 x^2+4 y^2 \\
\Rightarrow 3 x^2+4 y^2=0\end{array}$

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MCQ 2152 Marks

The equation of the locus of a point whose distance from (a, 0) is equal to its distance from Y-axis is

A
$y^2-2 a x= a ^2$
B
$y^2+2 a x+a^2=0$
✓
$y^2-2 a x+a^2=0$
D
$y^2+2 a x=a^2$

Answer

Correct option: C.

$y^2-2 a x+a^2=0$

(C)
Let $P ( h , k )$ be any point on the locus.
distance of P from Y -axis $= h$
According to the given condition,
$\begin{array}{l}\sqrt{(h-a)^2+(k-0)^2}=h \\
\Rightarrow(h-a)^2+(k-0)^2=h^2 \\
\Rightarrow h^2+a^2-2 ah+k^2=h^2\end{array}$
Hence, locus of $( h , k )$ is $y^2-2 a x+ a ^2-0$.

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MCQ 2162 Marks

Locus of a point, distance of which from origin is five times its distance from (3,-2), is

✓
$24 x^2+24 y^2-150 x+100 y+325=0$
B
$3 x^2+2 y^2+6 x+12 y-6=0$
C
$2 x^2+3 y^2+12 x+6 y-24=0$
D
$6 x^2+6 y^2-3 x+6 y+7=0$

Answer

Correct option: A.

$24 x^2+24 y^2-150 x+100 y+325=0$

(A)
Let $P (x, y)$ be any point on the locus.
Let $A \equiv(3,-2), O \equiv(0,0)$
We have, $OP =5 AP$
$\Rightarrow \sqrt{(x-0)^2+(y-0)^2}=5 \sqrt{(x-3)^2+(y+2)^2}$
Squaring both sides, we get
$(x-0)^2+(y-0)^2=25\left[(x-3)^2+(y+2)^2\right]$
$\Rightarrow 24 x^2+24 y^2-150 x+100 y+325=0$

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MCQ 2172 Marks

For two points A (2, 3) and B(- 2, 5) the locus of point P such that PA : PB :: 2 : 1 is

A
$x^2+3 y^2+20 x-34 y+103=0$
B
$3 x^2+3 y^2+18 x+34 y+103=0$
✓
$3 x^2+3 y^2+20 x-34 y+103=0$
D
$x^2+y^2+20 x+24 y+105=0$

Answer

Correct option: C.

$3 x^2+3 y^2+20 x-34 y+103=0$

(C)
Let $P \equiv(x, y)$
Given, $\frac{ PA }{ PB }=2$
$\therefore \quad PA =2 PB$
$\therefore \quad \sqrt{(x-2)^2+(y-3)^2}=2 \sqrt{(x+2)^2+(y-5)^2}$
Squaring both sides, we get
$(x-2)^2+(y-3)^2=4\left[(x+2)^2+(y-5)^2\right]$
$\Rightarrow 3 x^2+3 y^2+20 x-34 y+103=0$

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MCQ 2182 Marks

If the distance of any point P from the points A(a + b, a - b) and B(a - b, a + b) are equal, then the locus of P is

✓
x - y = 0
B
ax + by = 0
C
bx - ay = 0
D
x + y = 0

Answer

Correct option: A.

x - y = 0

(A)
Let the point P bc $(x, y)$.
Given, $PA = PB$
$\Rightarrow( PA )^2=( PB )^2$
$\Rightarrow\{x-( a + b )\}^2+\{y-( a - b )\}^2$
$=\{x-( a - b )\}^2+\{y-( a + b )\}^2$
$\Rightarrow 2 x(-a-b+a-b)+2 y(-a+b+a+b)=0$
$\Rightarrow 2 x(-2 b)+2 y(2 b)=0$
$\Rightarrow-x+y=0$
$\Rightarrow x-y=0$

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MCQ 2192 Marks

If A(- 5, 2) and B(- 1, 5) are two points, then the equation of perpendicular bisector of segment AB is

A
8x + 6y - 3 = 0
✓
8x + 6y + 3 = 0
C
8x - 6y + 3 = 0
D
8x - 6y - 3 = 0

Answer

Correct option: B.

8x + 6y + 3 = 0

(B)
Let $P (x, y)$ be any point on perpendicular bisector of AB .
$\begin{array}{l}\therefore \quad P A=P B \\\quad \Rightarrow \sqrt{(x+5)^2+(y-2)^2}=\sqrt{(x+1)^2+(y-5)^2}\end{array}$
Squaring both sides, we get
$\begin{array}{l}(x+5)^2+(y-2)^2=(x+1)^2+(y-5)^2 \\\Rightarrow 8 x+6 y+3=0\end{array}$

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MCQ 2202 Marks

Equation of locus of a point which is equidistant from (2,3) and (-4,5) is

A
3x - 7y = 0
B
x - 3y = - 5
C
x - 3y = 5
✓
3x - y = - 7

Answer

Correct option: D.

3x - y = - 7

(D)
$(x-2)^2+(y-3)^2=(x+4)^2+(y-5)^2$
$\Rightarrow x^2-4 x+4+y^2-6 y+9$
$=x^2+8 x+16+y^2-10 y+25$
$\Rightarrow-12 x+4 y=28$
$\Rightarrow 3 x-y=-7$

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Maths MCQ