# CBSE Class 11th Math 11 – Conic Sections MCQs

#### If y = 2x is a chord of the circle x2 + y2 - 10x = 0, then the equation of the circle with the chord as diameter is ____.

Correct! Wrong!

The chord y = 2x and the circle intersect the circle x2 + y2 - 10x = 0. Substitute y = 2x in the circle equation 4x2 + x2 - 10x = 0 = 5x(x - 2) = 0. x = 0 and x = 2, The chords end points are (0, 0) and (2, 4). Then, the equation of the circle with these points as end points is, (x - 0)(x - 2) + (y - 0)(y - 4) = 0 x2 + y2 - 2x - 4y = 0

#### The focal distance of a point on the parabola y2 = 16x is 5. The abcissa of this point is _____.

Correct! Wrong!

Given y2 = 16x. Therefore, 4a = 16? a = 4. The focal distance of any point (x, y) on y2 = 4 ax is x + a. Therefore, the focal distance is x + 4. Therefore, x + 4 = 5? x = 1. Therefore, The abcissa of the given point is 1.

#### The equation of the directrix of x2 = -8y is _____.

Correct! Wrong!

The given equation is x2 = -8y?a = + 8/4= + 2 The equation of the directrix is y = a (i.e) y = 2

#### The length of the latus rectum of x2 = 9y is equal to _______.

Correct! Wrong!

The given equation x2 = 9y is of the form x2 = 4 ay. i.e. 4 a = 9? a =9/4. So, this is an upward parabola. The length of the latus rectum is 4 a = 4? 9/4 = 9 units

#### The general equation of the ellipse is (x2/a2) + (y2b2) = 1. If a2 > b2, then we get a horizontal ellipse.

Correct! Wrong!

If a2 > b2, then we get a horizontal ellipse.

#### The length of the transverse axis is the distance between the ____.

Correct! Wrong!

The length of the transverse axis is the distance between the two vertices.

#### A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror whose equation is (y - 2)2 = 4(x + 1). Then, after reflection, the ray must pass through the point ___________.

Correct! Wrong!

The equation of the axis of parabola is y - 2 = 0, and which is parallel to the x-axis, is the axis of the mirror. Any ray that is parallel to the axis, should pass through the focus after reflection. Shifting the origin to (-1, 2), if (x, y) transforms to (X, Y), (y - 2)2= 4(x + 1) becomes x = X - 1 and y = Y + 2 Y2 = 4X comparing with standard form y2 = 4ax, a = 1 Focus with respect to new axes (X, Y) is (1, 0). With respect to original axis, x = 1 - 1 = 0, y = 0 + 2 Therefore, the focus is (0, 2).

#### The foci of an ellipse always ______.

Correct! Wrong!

The foci of an ellipse always lie on the major axis.

#### The co-ordinates of the focus and vertex of the parabola y2 = -4x are ____.

Correct! Wrong!

The given equation is y2 = -4x? 4 a = 4, i.e. a = 1 This is a left handed parabola. It's focus is (-a, 0) Therefore, F (-1, 0). It's vertex is (0, 0)

#### The equation of the parabola with vertex at the origin, passing through the point (2, 3) and symmetric with respect to Y-axis is ______.

Correct! Wrong!

The vertex is (0, 0) and symmetric about Y-axis. Therefore, Its equation is x2 = ± 4 ay (5, 3) lies on the parabola. (2, 3) lies in the first quadrant. Therefore, It is an upward parabola. Therefore, assume the equation as x12 = 4 ay1 (5, 3) lies on x12 = 4 ay1? 22 = 4 × a × 3 = 12a 4 = 12 a?a = 4/12 = 13. Therefore, the required equation is x2 = 4? 1/3? y? 3x2=4 y.

Correct! Wrong!

Hyperbola

#### A parabolic reflector is 10 cm in diameter. It's depth is 2.5 cm. The focus will be _______.

Correct! Wrong!

Let (0, 0) be the origin. Given : diameter PQ = 10 cm Therefore, PN = NQ = PQ/2 = 5 cm ON = 2.5 cm OX is the X axis. OY represents the Y axis. Equation of the reflector is, y2 = 4 ax. P lies on this parabola. Co-ordinates of P are (2.5, 5). Therefore, (5)2= 4 × 5 × a? 25 = 10 a? a = 5/2. Therefore, the equation of the reflector is y2 = 4 × 5/2 × x? y2 = 10x Focus F is at (2.5, 0) which is N.

#### The points on the parabola y2 = 18x, where the ordinate is equal to three times the abscissa are _________________.

Correct! Wrong!

Let (p, q ) be the point on the parabola. Then, this point should satisfy the parabola equation. q2 = 18p, since the ordinate is 3 times the abscissa, q = 3p. (3p)2= 18p i.e. 9 p2 - 18p = 0 => 9p(p - 2) = 0 P = 0 or p = 2. The corresponding values of q are 0 and 6. Hence, the points are (0, 0) and (2, 6).

#### The centre of the circle drawn on the intercepts between the axes made by the line 3x + y = 12 as diameter is ____.

Correct! Wrong!

Intercepts made by the line 3x + y = 12 (3x/12) +(y/12) = 1 The x-intercept is 4 and y intercept is 12. it means that the circle passes through (4, 0) and (0,12). These are the end points of the diameter of the circle. Therefore, the centre of the circle is [(4+0)/2, (12+0/2)] = (2, 6)

#### (x2/9)+(y2/25) = 1 represents a vertical ellipse.

Correct! Wrong!

Given (x2/9)+(y2/25) = 1 Comparing with the standard equation, we get, a2 = 9? a = 3 and b2 =25? b = 5 As 3 < 5, a < b Therefore, it represents a vertical ellipse.

#### The equation of the directrix of the parabola y2 = 16x is _______.

Correct! Wrong!

The given equation is, y2 = 16x is of the form y2 = 4 ax ? 4a = 1+ (i.e.) a = 4 This is a right handed parabola. The equation of the directrix is x + a = 0? x + 4 = 0 ? x = - 4

#### The equation of the parabola with focus F (4, 0) and directrix x = -4 is given by ______.

Correct! Wrong!

The focus is (4, 0).Therefore, It lies on the X-axis. As y = 0, X-axis is the axis of the parabola. (4, 0) is to the right of the origin. Hence, it is a right handed parabola. Let the required equation be y2 = 4 ax. Focus is F (4, 0). Therefore, a = 4. Therefore, the equation is y2 = 4 × 4 × x = 16x

#### The area of the circle is 154 square units. The equation of 2 of its diameters are x + y = 6 and x + 2y = 4. Then, the equation of the circle is ____.

Correct! Wrong!

The centre of the circle is the intersection of x + y = 6 and x + 2y = 4: y = -2, then x = 8. Area of the circle is 154 square units: π r2 = 154, assuming π = 22/7. 22/7 r2 = 154 r2= 154 (7/22) = 49 and r = 7 The equation of the circle is (x - 8)2 + (y + 2)2 = 72 x2 - 16x + 64 + y2 +2y + 4 = 49 x2 + y2 - 16x + 2y + 19 = 0

#### If one end point of the diameter of the circle x2 + y2 - 8x - 4y + c = 0 is (-3, 2), then the other end of the diameter of the circle is ____.

Correct! Wrong!

The equation of a circle x2 + y2 + 2gx + 2fy + c = 0. (-g, -f) is the centre. Comparing with the equation x2 + y2 - 8x - 4y + c = 0, the centre of the circle is (4, 2). One end point of the diameter is (-3, 2). Let the other end be (x, y). (x –3)/2 = 4 ⇒ x = 11 similarly, (y+2)/2 = 2 ⇒ y = 2 Therefore, the other end point is (11, 2).

#### A circle of radius 5 units has its centre on the positive x-axis. If the circle passes through the point (2, 3), then the centre of the circle is ____.

Correct! Wrong!

The centre of the required circle lies on x-axis means that the centre of the circle is (x, 0). It is given that the circle passes through the point (2, 3). Hence, (x - 2)2 + (0 - 3)2 = 52 x2 - 4x + 4 + 9 = 25 x2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 The centre is (6, 0). (x being positive, x ≠ -2)

#### The co-ordinates of the foci of an ellipse are (4, 0), (-4, 0). It represents a ______.

Correct! Wrong!

The co-ordinates of foci are (4, 0) and (-4, 0) The major axis is X-axis, as y = 0. Therefore, it represents a horizontal ellipse.

#### Choose the false statement.

Correct! Wrong!

The false statement is y2 = -4 ax represents the right handed parabola.

#### The equation of the parabola with vertex at (0, 0) and focus at ((-5/2), 0) is _____________.

Correct! Wrong!

Focus lies on the negative of x-axis. The parabola is opening to the left and is of the form y2 = -4ax. Here, a = -5/2. Therefore, the equation of the parabola is y2 = -4 ×(5/2) × x. => y2 = -10x

#### The equation of the circle having radius 3 units and which touches the y-axis at the origin and lies in the 1st and 4thquadrant is ____.

Correct! Wrong!

Since the circle touches y-axis at the origin and radius is of 3 units, the centre of the circle is (3, 0). The equation of the circle is, (x - 3)2 + y2 = 32 x2 - 6x + y2 = 0

#### The eccentricity of the ellipse is ______.

Correct! Wrong!

For an ellipse, e = C/a Because, C < a, we have e = C/a < 1? 0 < e < 1

CBSE Class 11th Math 11 - Conic Sections MCQs
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