MCQ - MATHS STD 11 Science Questions - Vidyadip

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21 questions · auto-graded multiple-choice test.

MCQ 11 Mark

The line $2x - y + 4 = 0$ cuts the parabola $y^2 = 8x$ in P and Q. The mid-point of PQ is

A
(1, 2)
B
(1, -2)
✓
(-1, 2)
D
(-1, -2)

Answer

Correct option: C.

(-1, 2)

(-1, 2)

Solution:
Let the coordinates of P and Q be (at$_1^2, 2$at$_1$) and (at$_2^2$, 2at$_2$), respectively.
$\text{Slope of PQ }=\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}...(1)$
But, the slope of PQ is equal to the slope of $2x - y + 4 = 0$.
$\therefore\ \text{Slope of PQ}=\frac{-2}{-1}=2$
From (1),
$\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}=2\ ...(2)$
Putting 4a = 8,
a = 2
$\therefore$ Focus of the given parabola = (a, 0) = (2, 0)
Using equation (2):
$\frac{4(\text{t}_2-\text{t}_1)}{2(\text{t}_2^2-\text{t}_1^2)}=2$
$\frac{(\text{t}_2-\text{t}_2)}{(\text{t}_2^2-\text{t}_1^2)}=1$
$\Rightarrow\ \text{t}_1+\text{t}_2=1$
As, points $P$ and $Q$ lie on $2 x-y+4=0$
$\Rightarrow P\left(a t_1{ }^2, 2 a t_1\right) \text { or } P\left(2 t_1{ }^2, 4 t_1\right) \text { lie on line } 2 x-y+4=0$
$\Rightarrow 2\left(2 t_1{ }^2\right)-\left(4 t_1\right)+4=0$
$\Rightarrow t_1{ }^2-t_1+1=0 \ldots(3)$
Also, $Q\left(a t_2{ }^2, 2 a t_2\right)$ or $P\left(2 t_2{ }^2, 4 t_2\right)$ lie on line $2 x-y+4=0$
$\Rightarrow 2\left(2 t_2^2\right)-\left(4 t_2\right)+4=0$
$\Rightarrow t_2^2-t_2+1=0 \ldots(4)$
Adding (3) and (4), we get,
$\Rightarrow t_1^2-t_1+1+t_2^2-t_2+1=0$
$\Rightarrow\left(t_1^2+t_2^2\right)-\left(t_1+t_2\right)+2=0$
$\Rightarrow\left(t_1^2+t_2^2\right)-1+2=0\left[t_1+t_2=1, \text { proved above }\right]$
$\Rightarrow\left(t_1^2+t_2^2\right)=-1$
Let $\left(x_1, y_1\right)$ be the mid-point of PQ.
Then, we have:
$\text{y}_1=\frac{2\text{at}_2+2\text{at}_1}{2}=2(\text{t}_1+\text{t}_2)=2$
And, $\text{x}_1=\frac{\text{at}_1^2+\text{at}_2^2}{2}=\text{t}_1^2+\text{t}_2^2=-1$
$\Rightarrow\ (\text{x}_1, \text{y}_1)=(-1, 2)$

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MCQ 21 Mark

The equation of the parabola whose vertex is $(a, 0)$ and the directrix has the equation $x + y = 3a$, is

A
$x^2+y^2+2 x y+6 a x+10 a y+7 a^2=0$
✓
$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$
C
$x^2-2 x y+y^2-6 a x+10 a y-7 a^2=0$
D
None of these

Answer

Correct option: B.

$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$

$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$

Solution:
Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.
The slope of the line perpendicular to x + y = 3a is 1.
The axis of the parabola is perpendicular to the directrix and passes through the vertex.
$\therefore$ Equation of the axis of the parabola = y − 0 = 1(x - a) ...(1)
Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.
Let the intersection point be K.
Therefore, the coordinates of K are (2a, a)
The vertex is the mid-point of the segment joining K and the focus (h, k).
$\therefore\ \text{a}=\frac{2\text{a+h}}{2},\ 0=\frac{\text{a+k}}{2}$
h = 0, k = -a
Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y= 3a.

Draw PM perpendicular to x + y = 3a.
Then, we have:
$SP = PM$
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x}-0)^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ 2\text{x}^2+2\text{y}^2+2\text{a}^2+4\text{ay}=\text{x}^2+\text{y}^2+9\text{a}^2+2\text{xy}-6\text{ax}-6\text{ay}$
$\Rightarrow\ \text{x}^2+\text{y}^2-7\text{a}^2+10\text{ay}+6\text{ax}=0$

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MCQ 31 Mark

The vertex of the parabola $(y + a)^2 = 8a (x - a)$ is

A
(-a, -a)
✓
(a, -a)
C
(-a, a)
D
None of these

Answer

Correct option: B.

(a, -a)

(a, a)

Solution:
Given:
The equation of the parabola is $(y + a)^2 = 8a (x - a)$.
Putting $X = x - a, Y = y + a$
$Y^2 = 8aX$
Vertex = $(X = 0, Y = 0) = (x - a = 0, y + a = 0) = (x = a, y = -a)$
Hence, the vertex is at (a, a).

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MCQ 41 Mark

If V and S are respectively the vertex and focus of the parabola $y^2 + 6y + 2x + 5 = 0$, then SV =

A
$2$
✓
$\frac{1}{2}$
C
$1$
D
None of these

Answer

Correct option: B.

$\frac{1}{2}$

$\frac{1}{2}$

Solution:
Given:
The vertex and the focus of a parabola are V and S , respectively.
The given equation of parabola can be rewritten as follows:
$(y+3)^2-9+5+2 x=0$
$\Rightarrow(y+3)^2+2 x=4$
$\Rightarrow(y+3)^2=4-2 x$
$\Rightarrow(y+3)^2=-2(x-2)$
$\text { Let } Y=y+3, x=x-2$
Then, the equation of parabola becomes $Y^2=-2 X$.
$\text { Vertex }=(X=0, Y=0)=(x-2=0, y+3=0)=(x=2, y=-3)$
Comparing with $y^2=4 a x$ :
$4\text{a} = 2 \Rightarrow \text{a} =\frac{1}{2}$
Focus $=\ \Big(\text{X}=\frac{-1}{2}, \text{Y}=0\Big)=\Big(\text{x}-2=\frac{-1}{2},\text{y}+3=0\Big)=\Big(\text{x}=\frac{3}{2},\text{y}=-3\Big)$
$\Rightarrow\ \text{SV}=\sqrt{\Big(2-\frac{3}{2}\Big)^2+(-3+3)^2}=\frac{1}{2}$

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MCQ 51 Mark

The coordinates of the focus of the parabola $y^2 - x - 2y + 2 = 0$ are

✓
$\Big(\frac{5}{4}, 1\Big)$
B
$\Big(\frac{1}{4}, 0\Big)$
C
$(1, 1)$
D
None of these

Answer

Correct option: A.

$\Big(\frac{5}{4}, 1\Big)$

$\Big(\frac{5}{4}, 1\Big)$

Solution:
Given:
The equation of the parabola is $y^2 - x - 2y + 2 = 0$.
$\Rightarrow (y - 1) - 1 = (x - 2)$
$(y - 1) = x - 1$
Let $X = x - 1, Y = y - 1$
$Y = X$
Comparing with Y = 4aX:
$\text{a}=\frac{1}{4}$
Focus=
$(\text{X} = \text{a}, \text{Y} = 0) = (\text{X} = \frac{1}{4}, \text{Y} = 0) = (\text{x} = \frac{1}{4}+ 1, \text{y} = 1) = (\text{x} = \frac{5}{4}, \text{y} = 1)$
Hence, the focus is at $\Big(\frac{5}{4}, 1\Big)$

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MCQ 61 Mark

The focus of the parabola $y = 2x^2 + x$ is

A
$(0, 0)$
B
$\Big(\frac{1}{2}, \frac{1}{4}\Big)$
✓
$\Big(-\frac{1}{4},0\Big)$
D
$\Big(-\frac{1}{4}, \frac{1}{8}\Big)$

Answer

Correct option: C.

$\Big(-\frac{1}{4},0\Big)$

$\Big(-\frac{1}{4},0\Big)$

Solution:
Given:
Equation of the parabola = $y = 2x^2 + x$
$\Rightarrow\ \text{x}^2+\frac{\text{x}}{2}=\frac{\text{y}}{2}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{\text{y}}{2}+\frac{1}{16}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{8\text{y}+1}{16}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{1}{2}(\text{y}+\frac{1}{8})$
$\text{Let }\text{X}=\text{x}+\frac{1}{4},\text{Y}=\text{y}+\frac{1}{8}$
$\therefore\ \text{X}^2=\frac{1}{2}\text{Y}$
Comparing with X = 4aY
$\text{a}=\frac{1}{8}$
Focus $=(\text{X}=0,\ \text{Y}=\text{a})=\Big(\text{x}=\frac{-1}{4},\text{y}=0\Big)$
Hence, the focus is at $\Big(-\frac{1}{4},0\Big).$

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MCQ 71 Mark

The vertex of the parabola $x^2 + 8x + 12y + 4 = 0$ is

✓
(-4, 1)
B
(4, -1)
C
(-4, -1)
D
(4, 1)

Answer

Correct option: A.

(-4, 1)

(-4, 1)

Solution:
Given:
$x^2+8 x+12 y+4=0$
$\Rightarrow(x+4)^2-16+12 y+4=0$
$\Rightarrow(x+4)^2+12 y-12=0$
$\Rightarrow(x+4)^2=-12(y-1)$
Let $X=x+4, Y=y-1$
$X^2=-12 Y$
Vertex $=(X=0, Y=0)=(x+4=0, y-1=0)=(x=-4, y=1)$
Hence, the vertex is at $(-4,1)$.

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MCQ 81 Mark

If the focus of a parabola is (-2, 1) and the directrix has the equation x + y = 3, then its vertex is

A
$ (0, 3)$
B
$\Big(−1,\ \frac{1}{2}\Big)$
✓
$(−1, 2)$
D
$ (2, −1)$

Answer

Correct option: C.

$(−1, 2)$

Given:
The focus S is at (-2, 1) and the directrix is the line x + y - 3 = 0.
The slope of the line perpendicular to x + y - 3 = 0 is 1.
The axis of the parabola is perpendicular to the directrix and passes through the focus.
$\therefore$ Equation of the axis of the parabola = y - 1 = 1(x + 2) ...(1)
Intersection point of the directrix and the axis is the intersection point of (1) and x + y - 3 = 0.
Let the intersection point be K.
Therefore, the coordinates of K will be (0, 3).
Let (h, k) be the coordinates of the vertex, which is the mid-point of the segment joining K and the focus.
$\therefore\ \text{h}=\frac{0-2}{2},\ \text{k}=\frac{3+1}{2}$
h = -1, k = 2
Hence, the coordinates of the vertex are $(−1, 2).$

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MCQ 91 Mark

The vertex of the parabola $(y - 2)^2 = 16 (x - 1)$ is

✓
(1, 2)
B
(-1, 2)
C
(1, -2)
D
(2, 1)

Answer

Correct option: A.

(1, 2)

(1, 2)

Solution:
Given:
$(y-2)^2=16(x-1)$
$\text { Let } X=x-1, Y=y-2$
$\therefore Y^2=16 X$
$\text { Vertex }=(X=0, Y=0)=(x-1=0, y-2=0)=(x=1, y=2)$
Hence, the vertex is at $(1,2)$.

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MCQ 101 Mark

In the parabola $y^2 = 4ax$, the length of the chord passing through the vertex and inclined to the axis at $\frac{\pi}{4}$ is

✓
$4\sqrt2\text{a}$
B
$2\sqrt2\text{a}$
C
$\sqrt2\text{a}$
D
None of these

Answer

Correct option: A.

$4\sqrt2\text{a}$

$4\sqrt2\text{a}$

Solution:

Let $O P$ be the chord.
Let the coordinates of $P$ be $\left(\mathrm{x}_1, \mathrm{y}_1\right)$.
From the figure, we have:
$\mathrm{OP}^2=\mathrm{x}_1^2+\mathrm{y}_1^2 \ldots(1)$
And, $\tan \frac{\pi}{4}=\frac{\mathrm{y}_1}{\mathrm{x}_1}$
$\Rightarrow x_1=y_1 \ldots(2)$
Also, $\left(x_1, y_1\right)$ lies on the parabola.
$\therefore y_1^2=4 a x_1 \ldots(3)$
Using (2) and (3):
$\mathrm{x}_1^2=4 a \mathrm{x}_1 \Rightarrow \mathrm{x}_1=4 \mathrm{a}$
From (4), (1) and (2), we have:
$O P^2=(4 a)^2+(4 a)^2=32 a^2$
$\Rightarrow O P=4 \sqrt{2} a$
Therefore, the length of the chord is $4 \sqrt{2} \mathrm{a}$ a units.

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MCQ 111 Mark

If the coordinates of the vertex and the focus of a parabola are (-1, 1) and (2, 3) respectively, then the equation of its directrix is

✓
3x + 2y + 14 = 0
B
3x + 2y - 25 = 0
C
2x - 3y + 10 = 0
D
None of these

Answer

Correct option: A.

3x + 2y + 14 = 0

Given:
The vertex and the focus of a parabola are (-1, 1) and (2, 3), respectively.
$\therefore$ Slope of the axis of the parabola $=\frac{3-1}{2+1}=\frac{2}{3}$
Slope of the directrix $=\ \frac{-3}{2}$
Let the directrix intersect the axis at K (r, s).
$\therefore\ \frac{\text{r+2}}{2}=-1,\ \frac{\text{s}+3}{2}=1$
$\Rightarrow\ \text{r}=-4,\ \text{s}=-1$
Equation of the directrix:
$(\text{y}+1)=\frac{-3}{2}(\text{x}+4)$
$\Rightarrow\ 3\text{x}+2\text{y}+14=0$

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MCQ 121 Mark

Which of the following points lie on the parabola $x^2=4 a y ?$

A
$x=a t^2, y=2 a t$
B
$x=2 a t, y=a t^2$
C
$x=2 a t^2, y=a t$
✓
$x=2 a t, y=a t^2$

Answer

Correct option: D.

$x=2 a t, y=a t^2$

$x=2 a t, y=a t^2$

Solution:
Substituting $x=2 a t, y=a t^2$ in the given equation:
$(2 a t)^2=4 a\left(a t^2\right)$
$\Rightarrow 4 a^2 t^2=4 a^2 t^2$
Hence, (2at, at $\left.{ }^2\right)$ lies on the parabola $x^2=4 a y$.

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MCQ 131 Mark

The parametric equations of a parabola are $x = t^2 + 1, y = 2t + 1$. The cartesian equation of its directrix is

✓
x = 0
B
x + 1 = 0
C
y = 0
D
None of these

Answer

Correct option: A.

x = 0

x = 0

Solution:
Given:
$x=t^2+1 \ldots(1)$
$y=2 t+1 \ldots(2)$
From (1) and (2):
$x=\left(\frac{y-1}{2}\right)^2+1$
On simplifying:
$(y-1)^2=4(x-1)$
Let $Y=y-1$ and $X=x-1$
$\therefore \mathrm{Y}^2=4 \mathrm{X}$
Comparing it with $y^2=4 a x$ :
$a=1$
Therefore, the equation of the directrix is $X=-$ a, i.e. $x-1=-1 \Rightarrow x=0$

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MCQ 141 Mark

The equation of the directrix of the parabola whose vertex and focus are (1, 4) and (2, 6) respectively is

✓
x + 2y = 4
B
x - y = 3
C
2x + y = 5
D
x + 3y = 8

Answer

Correct option: A.

x + 2y = 4

Given:
The vertex and the focus of a parabola are (1, 4) and (2, 6), respectively.
$\therefore$ Slope of the axis of the parabola $= \frac{6-4}{2-1}=2$
Slope of the directrix $=\ \frac{-1}{2}$
Let the directrix intersect the axis at K (r, s).
$\therefore\ \frac{\text{r}+2}{2}=1,\ \frac{\text{s}+6}{2}=4$
$\Rightarrow\ \text{r}=0,\ \text{s}=2$
Equation of the directrix:
$(\text{y}-2)=\frac{-1}{2}(\text{x}-0)$
⇒ x + 2y = 4

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MCQ 151 Mark

The length of the latus-rectum of the parabola $y^2 + 8x − 2y + 17 = 0$ is

A
2
B
4
✓
8
D
16

Answer

Correct option: C.

8

8

Solution:
$y^2+8 x-2 y+17=0$
$\Rightarrow(y-1)^2-1+8 x+17=0$
$\Rightarrow(y-1)^2+8 x+16=0$
$\Rightarrow(y-1)^2=-8(x+2)$
Let $X=x+2, Y=y-1$
$\therefore \mathrm{Y}^2=-8 \mathrm{X}$
Comparing with $y^2=4 a x$ :
$a=2$
Length of the latus rectum $=4 a=8$ units

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MCQ 161 Mark

The length of the latus-rectum of the parabola $x^2 - 4x - 8y + 12 = 0$ is

A
4
B
6
✓
8
D
10

Answer

Correct option: C.

8

8

Solution:
Given:
$x^2 - 4x - 8y + 12 = 0$
$(x - 2)^2 - 8y + 8 = 0$
$(x - 2)^2 = 8y - 8 = 8(y - 1)$
Let $X = x - 2, Y = y - 1$
$\therefore X^2 = 8Y$
$\therefore$ Length of the latus rectum = $4a = 8$ units

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MCQ 171 Mark

The equation $16x^2 + y^2 + 8xy - 74x - 78y + 212 = 0$ represents

A
A circle
✓
A parabola
C
An ellipse
D
A hyperbola

Answer

Correct option: B.

A parabola

a parabola

Solution:
Comparing the given equation with $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$, we get:
$a = 16, b = 1, h = 4$
We have: $h^2 = 16 = ab$
Thus, the given equation represents a parabola.

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MCQ 181 Mark

The directrix of the parabola $x^2 - 4x - 8y + 12 = 0$ is

A
y = 0
B
x = 1
✓
y = -1
D
x = -1

Answer

Correct option: C.

y = -1

y = -1

Solution:
Given:
$x^2-4 x-8 y+12=0$
$\Rightarrow(x-2)^2-4-8 y+12=0$
$\Rightarrow(x-2)^2=8 y-8$
$\Rightarrow(x-2)^2=8(y-1) \backslash$
Putting $X=x-2, Y=y-1$ :
$X^2=8 Y$
Comparing with $\mathrm{X}^2=4 \mathrm{a} \mathrm{Y}$ :
$a=2$
Equation of the directrix:
$Y=-a$
$\Rightarrow Y=-2$
$\Rightarrow y-1=-2$
$\Rightarrow y=-2+1$
$\Rightarrow y=-1$

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MCQ 191 Mark

The equation of the parabola whose focus is $(1,-1)$ and the directrix is $x+y+7=0$ is

A
$x^2+y^2-2 x y-18 x-10 y=0$
B
$x^2-18 x-10 y-45=0$
C
$x^2+y^2-18 x-10 y-45=0$
✓
$x^2+y^2-2 x y-18 x-10 y-45=0$

Answer

Correct option: D.

$x^2+y^2-2 x y-18 x-10 y-45=0$

$x^2+y^2-2 x y-18 x-10 y-45=0$

Solution:
Let $P (x, y)$ be any point on the parabola whose focus is $S (1, -1)$ and the directrix is $x + y+ 7 = 0$.

Draw PM perpendicular to $x + y + 7 = 0$.
Then, we have:
$SP = PM$
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x} - 1)^2+ (\text{y} + 1)^2= \Big(\frac{\text{x+y+7}}{\sqrt{1+1}}\Big)^2$
$\Rightarrow\ (\text{x} - 1)^2+ (\text{y} + 1)^2= \Big(\frac{\text{x+y+7}}{\sqrt{2}}\Big)^2$
$\Rightarrow\ 2(\text{x}^2+1-2\text{x}+\text{y}^2+1+2\text{y})\\ \ \ =\text{x}^2+\text{y}^2+49+2\text{xy}+14\text{y}+14\text{x}$
$\Rightarrow\ (2\text{x}^2+2-4\text{x}+2\text{y}^2+2+4\text{y})\\ \ \ =\text{x}^2+\text{y}^2+49+2\text{xy}+14\text{y}+14\text{x}$
$\Rightarrow\ \text{x}^2+\text{y}^2-45-10\text{y}-2\text{xy}-18\text{x}=0$
Hence, the required equation is $x^2 + y^2 - 2xy - 18x - 10y - 45 = 0$.

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MCQ 201 Mark

The equation of the parabola with focus $(0, 0)$ and directrix $x + y = 4$ is

✓
$x^2+y^2-2 x y+8 x+8 y-16=0$
B
$x^2+y^2-2 x y+8 x+8 y=0$
C
$x^2+y^2+8 x+8 y-16=0$
D
$x^2-y^2+8 x+8 y-16=0$

Answer

Correct option: A.

$x^2+y^2-2 x y+8 x+8 y-16=0$

$x^2+y^2-2 x y+8 x+8 y-16=0$

Solution:
Let $P (x, y)$ be any point on the parabola whose focus is $S (0, 0)$ and the directrix is $x + y = 4$.

Draw PM perpendicular to $x + y = 4$.
Then, we have:
$SP = PM$
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x}-0)^2+(\text{y}-0)^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$
$\Rightarrow 2x^2 + 2y^2 = x^2 + y^2 + 16 + 2xy - 8x - 8y$
$\Rightarrow x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$

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MCQ 211 Mark

The length of the latus-rectum of the parabola $4y^2 + 2x - 20y + 17 = 0$ is

✓
$3$$$
B
$6$
C
$\frac{1}{2}$
D
$9$

Answer

Correct option: A.

$3$$$

$\frac{1}{2}$

Solution:
Given:
$4y^2 + 2x - 20y + 17 = 0$
$\Rightarrow\ \text{y}^2+\frac{\text{x}}{2}-5\text{y}+\frac{17}{4}=0$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2+\frac{\text{x}}{2}-2=0$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=-1\Big(\frac{\text{x}}{2}-2\Big)$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=\frac{-1}{2}(\text{x}-4)$
$\text{Let }\text{X}=\text{x}-4,\ \text{Y}=\text{y}-\frac{5}{2}$
$\therefore\ \text{Y}^2=\frac{-\text{X}}{2}$
$\therefore$ Length of the latus rectum $=\ 4\text{a}=\frac{1}{2}$ units

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