**CBSE Class 11 Maths – MCQ and Online Tests – Unit 11 – Conic Sections**

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**CBSE Class 11 Maths – MCQ and Online Tests – Unit 11 – Conic Sections**

Question 1.

The radius of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?

(a) v57/4

(b) v77/4

(c) v77/2

(d) v87/4

## Answer

Answer: (c) v77/2

Hint:

Given, equation fo the of the circle is 4x² + 4y² – 8x + 12y – 25 = 0

? x² + y² – 8x/4 + 12y/4 – 25/4 = 0

? x² + y² – 2x + 3y – 25/4 = 0

Now, radius = v{(-2)² + (3)² – (-25/4)}

= v{4 + 9 + 25/4}

= v{13 + 25/4}

= v{(13×4 + 25)/4}

= v{(52 + 25)/4}

= v{77/4}

= v77/2

Question 2.

If (a, b) is the mid point of a chord passing through the vertex of the parabola y² = 4x, then

(a) a = 2b

(b) 2a = b

(c) a² = 2b

(d) 2a = b²

## Answer

Answer: (d) 2a = b²

Hint:

Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0)

Now, (x + 0)/2 = a

? x = 2a

and (y + 0)/2 = b

? y = 2b

Now, y² = 4x

? (2b)² = 4 × 2a

? 4b² = 8a

? b² = 2a

Question 3.

A man running a race course notes that the sum of the distances from the two flag posts from him is always 10 meter and the distance between the flag posts is 8 meter. The equation of posts traced by the man is

(a) x²/9 + y²/5 = 1

(b) x²/9 + y2 /25 = 1

(c) x²/5 + y²/9 = 1

(d) x²/25 + y²/9 = 1

## Answer

Answer: (d) x²/25 + y²/9 = 1

Hint:

From the question, it is clear that the path traced by the man is an ellipse having its foci at two posts.

Let the equation of the ellipse be

x²/a² + y²/b² = 1

It is given that the sum of the distances of the man from the two flag posts is 10 m

This means that the sum of focal distances of a point on the ellipse is 10 m

? PS + PS_{1} = 10

? 2a = 10

? a = 5

Again, given that the distance between the flag posts is 8 meters

? 2ae = 8

? ae = 4

Now, b² = a² (1 – e²)

? b² = a² – a² e²

? b² = a² – (ae)²

? b² = 5² – 4²

? b² = 25 – 16

? b² = 9

? b = 3

Hence, the equation of the path is x²/5² + y²/3² = 1

? x²/25 + y²/9 = 1

Question 4.

The center of the ellipse (x + y – 2)² /9 + (x – y)² /16 = 1 is

(a) (0, 0)

(b) (0, 1)

(c) (1, 0)

(d) (1, 1)

## Answer

Answer: (d) (1, 1)

Hint:

The center of the given ellipse is the point of intersection of the lines

x + y – 2 = 0 and x – y = 0

After solving, we get

x = 1, y = 1

So, the center of the ellipse is (1, 1)

Question 5.

The parametric coordinate of any point of the parabola y² = 4ax is

(a) (-at², -2at)

(b) (-at², 2at)

(c) (a sin²t, -2a sin t)

(d) (a sin t, -2a sin t)

## Answer

Answer: (c) (a sin²t, -2a sin t)

Hint:

The point (a sin²t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all

values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is

(a sin²t, -2a sin t)

Question 6.

The equation of parabola with vertex at origin the axis is along x-axis and passing through the point (2, 3) is

(a) y² = 9x

(b) y² = 9x/2

(c) y² = 2x

(b) y² = 2x/9

## Answer

Answer: (b) y² = 9x/2

Hint:

A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:

y² = 4ax ………….. 1

Given, point (2,3) lies on the parabola,

? 3² = 4a × 2

? 9 = 4a × 2

? 9/2 = 4a

From equation 1, we get

y² = (9/2)x

? y² = 9x/2

This is the required equation of the parabola.

Question 7.

The locus of the point from which the tangent to the circles x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0 are equal is given by the equation

(a) 8x + 19 = 0

(b) 8x – 19 = 0

(c) 4x – 19 = 0

(d) 4x + 19 = 0

## Answer

Answer: (b) 8x – 19 = 0

Hint:

Given equation of circles are x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0

Now, the required line is the radical axis of the two circles are

(x² + y² – 4) – (x² + y² – 8x + 15) = 0

? x² + y² – 4 – x² – y² + 8x – 15 = 0

? 8x – 19 = 0

Question 8.

The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0

(a) 7

(b) 8

(c) 9

(d) 10

## Answer

Answer: (a) 7

Hint:

The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/v(3² + 4²)

= {9 + 16 + 10}/v(9 + 16)

= 35/v25

= 35/5

= 7

Question 9.

In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is

(a) 4/5

(b) 1/v52

(c) 3/5

(d) 1/2

## Answer

Answer: (c) 3/5

Hint:

Given, distance between foci = 6

? 2ae = 6

? ae = 3

Again minor axis = 8

? 2b = 8

? b = 4

? b² = 16

? a² (1 – e²) = 16

? a² – a² e² = 16

? a² – (ae)² = 16

? a² – 3² = 16

? a² – 9 = 16

? a² = 9 + 16

? a² = 25

? a = 5

Now, ae = 3

? 5e = 3

? e = 3/5

So, the eccentricity is 3/5

Question 10.

If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is

(a) (x + 2)² + (y – 3)² = 3²

(b) (x – 2)² + (y + 3)² = 3²

(c) (x – 2)² + (y – 3)² = 3²

(d) (x + 2)² + (y + 3)² = 3²

## Answer

Answer: (c) (x – 2)² + (y – 3)² = 3²

Hint:

Radius of the circle = v{(2 – 0)² + (3 – 0)² – 2²}

= v(4 + 9 – 4)

= v9

= 3

So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Question 11.

The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is

(a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0

(b) 16x² + 9y² – 24xy – 144x + 8y – 224 = 0

(c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0

(d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

## Answer

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Hint:

Given focus S(3, 0)

and equation of directrix is: 3x + 4y = 1

? 3x + 4y – 1 = 0

Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix

Then, SP = PM

? SP² = PM²

? (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{v(3² + 4²)}²

? x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25

? 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x

? 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x

? 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0

? 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

This is the required equation of parabola.

Question 12.

The parametric representation (2 + t², 2t + 1) represents

(a) a parabola

(b) a hyperbola

(c) an ellipse

(d) a circle

## Answer

Answer: (a) a parabola

Hint:

Let x = 2 + t²

? x – 2 = t² ……….. 1

and y = 2t + 1

? y – 1 = 2t

? (y – 1)/2 = t

From equation 1, we get

x – 2 = {(y – 1)/2}²

? x – 2 = (y – 1)²/4

? (y – 1)² = 4(x – 2)

This represents the equation of a parabola.

Question 13.

The equation of a hyperbola with foci on the x-axis is

(a) x²/a² + y²/b² = 1

(b) x²/a² – y²/b² = 1

(c) x² + y² = (a² + b²)

(d) x² – y² = (a² + b²)

## Answer

Answer: (b) x²/a² – y²/b² = 1

Hint:

The equation of a hyperbola with foci on the x-axis is defined as

x²/a² – y²/b² = 1

Question 14.

The equation of parabola with vertex (-2, 1) and focus (-2, 4) is

(a) 10y = x² + 4x + 16

(b) 12y = x² + 4x + 16

(c) 12y = x² + 4x

(d) 12y = x² + 4x + 8

## Answer

Answer: (b) 12y = x² + 4x + 16

Hint:

Given, parabola having vertex is (-2, 1) and focus is (-2, 4)

As the vertex and focus share the same abscissa i.e. -2,

parabola axis of symmetry as x = -2

? x + 2 = 0

Hence, the equation of a parabola is of the type

(y – k) = a(x – h)² where (h, k) is vertex

Now, focus = (h, k + 1/4a)

Since, vertex is (-2, 1) and parabola passes through vertex

So, focus = (-2, 1 + 1/4a)

Now, 1 + 1/4a = 4

? 1/4a = 4 -1

? 1/4a = 3

? 4a = 1/3

? a = /1(3 × 4)

? a = 1/12

Now, equation of parabola is

(y – 1) = (1/12) × (x + 2)²

? 12(y – 1) = (x + 2)²

? 12y – 12 = x² + 4x + 4

? 12y = x² + 4x + 4 + 12

? 12y = x² + 4x + 16

This is the required equation of parabola.

Question 15.

If a parabolic reflector is 20 cm in diameter and 5 cm deep then the focus of parabolic reflector is

(a) (0 0)

(b) (0, 5)

(c) (5, 0)

(d) (5, 5)

## Answer

Answer: (c) (5, 0)

Hint:

given diameter of the parabola is 20 m.

The equation of parabola is y² = 4ax.

Since this parabola passes through the point A(5,10) then

10² = 4a×5

? 20a = 100

? a = 100/20

? a = 5

So focus of parabola is (a, 0) = (5, 0)

Question 16.

The line lx + my + n = 0 will touches the parabola y² = 4ax if

(a) ln = am²

(b) ln = am

(c) ln = a² m²

(d) ln = a² m

## Answer

Answer: (a) ln = am²

Hint:

Given, lx + my + n = 0

? my = -lx – n

? y = (-l/m)x + (-n/m)

This will touches the parabola y² = 4ax if

(-n/m) = a/(-l/m)

? (-n/m) = (-am/l)

? n/m = am/l

? ln = am²

Question 17.

The center of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?

(a) (2,-3)

(b) (-2,3)

(c) (-4,6)

(d) (4,-6)

## Answer

Answer: (a) (2,-3)

Hint:

Given, equation fo the of the circle is 4x² + 4y² – 8x + 12y – 25 = 0

? x² + y² – 8x/4 + 12y/4 – 25/4 = 0

? x² + y² – 2x + 3y – 25/4 = 0

Now, center = {-(-2), -3} = (2, -3)

Question 18.

A rod of length 12 CM moves with its and always touching the co-ordinate Axes. Then the equation of the locus of a point P on the road which is 3 cm from the end in contact with the x-axis is

(a) x²/81 + y²/9 = 1

(b) x²/9 + y²/81 = 1

(c) x²/169 + y²/9 = 1

(d) x²/9 + y²/169 = 1

## Answer

Answer: (a) x²/81 + y²/9 = 1

Hint:

Given a rod of length 12 cm moves with its ends always touching the coordinate axes.

Again given a point P on the rod, which is 3 cm from the end in contact with the x-axis.

It is shown in the figure.

Here AP = 3 cm, AB = 12

Now BP = AB – AP

? BP = 12 – 3

? BP = 9 cm

Again from figure,

?PAO = ?BPO = ? (since PQ || OA and are corresponding angles)

Now in ?BPO,

cos? = QP/BP

? cos? = x/9 …………. 1

Again in ?PAr,

sin? = PR/PA

? sin? = y/3 …….. 2

Now square equation 1 and 2 and then add them, we get

cos² ? + sin² ? = x²/81 + y²/9

? x²/81 + y²/9 = 1 (since cos² ? + sin² ? = 1 )

So, the equation of the locus of a point P is x²/81 + y²/9 = 1

Question 19.

At what point of the parabola x² = 9y is the abscissa three times that of ordinate

(a) (1, 1)

(b) (3, 1)

(c) (-3, 1)

(d) (-3, -3)

## Answer

Answer: (b) (3, 1)

Hint:

Given, parabola is x² = 9y

Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate.

So, h = 3k ……… 1

Since P(h, k) lies on the parabola

So, h² = 9k ……… 2

From equation 1 and 2, we get

(3k)² = 9k

? 9k² = 9k

? 9k² – 9k = 0

? 9k(k – 1) = 0

? k = 0, 1

When k = 0, h = 0

So k = 1

Now, from equation 1,

h = 3 × 1 = 3

So, the point is (3, 1)

Question 20.

The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is

(a) 0

(b) 1

(c) 2

(d) More than 2

## Answer

Answer: (b) 1

Hint:

Given point (1, 2) and equation of circle is x² + y² = 5

Now, x² + y² – 5 = 0

Put (1, 2) in this equation, we get

1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0

So, the point (1, 2) lies on the circle.

Hence, only one tangent can be drawn.

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