Practice AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections MCQ to identify your strengths and weak areas.
AP Inter 1st Year Maths Conic Sections MCQ
Circle
Question 1.
The diametes of a circle are along 2x + y – 7 = 0 and x + 3y – 11 = 0.Then the equation of this circle, which also passes through (5, 7) is ( 3 )
1) x2 + y2 – 4x – 6y – 16 = 0
2) x2 + y2 – 4x – 6y – 20 = 0
3) x2 + y2 – 4x – 6y – 12 = 0
4) x2 + y2 + 4x + 6y – 12 = 0
Answer:
3) x2 + y2 – 4x – 6y – 12 = 0
Explanation:
Point of intersection of 2x + y – 7 = 0, x + 3y – 11 = 0 is (2, 3).
2x + y – 7 = 0, x + 3y – 11 = 0 are diameters ⇒ Centre = C(2, 3)
Circle passes through A(5, 7) ⇒ Radius = CA = \(\sqrt{9+16}=\sqrt{25}\) = 5.
Equation of the circle is (x – 2)2 + (y – 3)2 = 52
∴ x2 + y2 – 4x – 6y – 12 = 0
Question 2.
Consider the circle x2 + y2 – 4x – 2y + c = 0 whose centre is A(2, 1). If the point P(10, 7) is such that the line segment PA meets the circle in 9 with PQ = 5, then c = ( 4 )
1) -15
2) 20
3) 30
4) -20
Answer:
4) -20
Explanation:
Given centre of the circle is x2 + y2 – 4x – 2y + c = 0 and centre A(2, 1) and P = (10, 7) PA meets the circle in Q with PQ = 5.
PA = PQ + QA
⇒ \(\frac{1}{2}\) = 5 + r
⇒ 10 = 5 + r
⇒ r = 5.
⇒ \(\frac{1}{2}\) = 5
⇒ \(\frac{1}{2}\) = 5
⇒ 5 – c = 25
⇒ c = -20
Question 3.
The abscissae of two points A and B are the roots of the equation x2 + 2ax – b2 = 0 and their ordinates are the roots of the equation y2 + 2py – q2 = 0. The radius of the circle with AB as diameter is
1) \(\sqrt{a^2+b^2+p^2+q^2}\)
2) \(\sqrt{a^2+p^2}\)
3) \(\sqrt{b^2+q^2}\)
4) \(\sqrt{a^2+b^2-p^2+q^2}\)
Answer:
1) \(\sqrt{a^2+b^2+p^2+q^2}\)
Explanation:
Let A = (x1, y1), B = (x2, y2)
x1, x2 are the roots of x2 + 2ax – b2 = 0
⇒ (x – x1) (x – x1) = 0
y1, y2 are the roots of y2 + 2py – q2 = 0
⇒ (x – y1) (x – y2) = 0
Equation of the circle with AB as diameter is
x2 + 2ax – b2 + y2 + 2py – q2 = 0
x2 + y2 + 2ax + 2py – b2 – q2 = 0
Radius of the circle = \(\\sqrt{a^2+p^2+b^2+q^2}=\sqrt{a^2+b^2+p^2+q^2}\)
Question 4.
The equation of the circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and of double its area is
1) x2 + y2 – 6x + 12y – 15 = 0
2) x2 + y2 – 6x + 12y – 30 = 0
3) x2 + y2 – 6x + 12y – 25 = 0
4) x2 + y2 – 6x + 12y – 20 = 0
Answer:
1) x2 + y2 – 6x + 12y – 15 = 0
Explanation:
Equation of the given circle is x2 + y2 – 6x + 12y + 15 = 0.
Radius r1 = \(\frac{1}{2}\)
Area of the given circle = 30π
Equation of the required circle is x2 + y2 – 6x + 12y + k = 0.
Radius r2 = \(\frac{1}{2}\)
Area of the required circle = (45 – k)π
Given that (45 – k)π = 2(30π) ⇒ k = – 15
Required circle is x2 + y2 – 6x + 12y – 15 = 0.
Question 5.
A square is inscribed in the circle x2 + y2 – 2x + 4y + 3 = 0 and its sides are parallel to the co-ordinate axes then one vertex of the square is
1) (1 + √2, -2)
2) (1 – √2, -2)
3) (1,- 2 + √2)
4) None
Answer:
4) None
Explanation:
Given x2 + y2 – 2x + 4y + 3 = 0
(x – 1)2 – 1 + (y + 2)2 – 4 + 3 = 0
⇒ (x – 1)2 + (y + 2)2 = 2
Centre C = (1, -2), Radius r = √2
Since diameter d = 2r = 2√2
Let ‘s’ is length of square, then d = s√2 ⇒ s√2 = 2√2 ⇒ s = 2
Half of side length is \(\frac{2}{2}\) = 1.
Since possible vertices are (1 + 1, -2 + 1) = (2, -1) ; (1 + 1, -2 – 1) = (2, -3)
(1 – 1, -2 + 1) = (0, -1) ; (1 – 1, -2 – 1) = (0, -3)
None of the option is correct.
Question 6.
The equation of the circle with centre (cos a, sin 1) and radius 1 is
1) x2 + y2 = 1
2) x2 + y2 – 2x cos a – 2y sin a = 0
3) x2 + y2 + 2x cos a + 2y sin a = 0
4) x2 + y2 – 2x cos a – 1 = 0
Answer:
2) x2 + y2 – 2x cos a – 2y sin a = 0
Explanation:
Equation of the circle with centre (cos a, sin a) and radius 1 is (x – cos a)2 + (y – sin a)2 = (1)2
⇒ x2 + y2 – 2x cos a – 2y sin a + cos2a + sin2a = 1
⇒ x2 + y2 – 2x cos a – 2y sin a = 0
Question 7.
If r1 and r2 are radii of the circles x2 + y2 – 4x – 8y – 44 = 0 and x2 + y2 + 6x + 8y – 96 = 0 respectively then r1 : r2 is
1) 8 : 11
2) 9 : 11
3) 16 : 21
4) 64 : 123
Answer:
1) 8 : 11
Explanation:
r1 = radius of x2 + y2 – 4x – 8y – 44 = 0
⇒ r1 = \(\sqrt{4+16+44}=\sqrt{64}\) = 8.
r2 = radius of x2 + y2 + 6x + 8y – 96 = 0
⇒ r2 = \(\sqrt{9+16+96}=\sqrt{121}\) = 11
∴ r1 : r2 = 8 : 11.
Question 8.
The radius of the circle touching the lines 3x – 4y + 5 = 0; 6x – 8y – 9 = 0 is
1) 1
2) \(\frac{23}{15}\)
3) \(\frac{20}{19}\)
4) \(\frac{19}{20}\)
Answer:
4) \(\frac{19}{20}\)
Explanation:
If r is the radius of the circle touching the lines 3x – 4y + 5 = 0; 6x – 8y – 9 = 0, then
2r = \(\frac{10+9}{\sqrt{36+64}}=\frac{19}{10}\)
⇒ r = \(\frac{19}{20}\)
Question 9.
The perimeter of the circle 3(x2 + y2) = 16 is
1) \(\frac{4}{\sqrt{3}}\)π
2) \(\frac{16}{\sqrt{3}}\)π
3) \(\frac{8}{\sqrt{3}}\)π
4) \(\frac{1}{\sqrt{3}}\)π
Answer:
3) \(\frac{8}{\sqrt{3}}\)π
Explanation:
Equation of the given circle is 3(x2 + y2) = 16
⇒ x2 + y2 = \(\frac{16}{3}\)
Perimeter of the circle = 2πr = 2π × \(\frac{4}{\sqrt{3}}=\frac{8}{\sqrt{3}}\)π
Question 10.
The radius of circle 3(x2 + y2) – 3x + 9y + 10 = 0
1) \(\sqrt{\frac{15}{2}}\)
2) \(\frac{5}{\sqrt{2}}\)
3) \(\sqrt{\frac{15}{3}}\)
4) Not defined
Answer:
4) Not defined
Explanation:
Equation of the given circle is 3(x2 + y2) – 3x + 9y + 10 = 0
⇒ x2 + y2 – x + 3y + \(\frac{10}{3}\) = 0
Radius of the circle = \(\frac{1}{2}\) Which is not defined.
Parabola
Question 1.
Parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at
1) (1, 0)
2) (0, 1)
3) (2, 0)
4) (0, 2)
Answer:
1) (1, 0)
Explanation:
S = (0, 0), Z = (2, 0)
⇒ A = \(\frac{S+Z}{2}\) = (1, 0)
Question 2.
If the focus of a parabola is (0, -3) and its directrix is y = 3 then its equation is
1) x2 = -12y
2) x2 = 12y
3) y2 =-12x
4) y2 = 12x
Answer:
1) x2 = -12y
Explanation:
Focus of the parabola is S = (0, -3)
Equation of the directrix is y = 3 ⇒ y – 3 = 0.
If P(x, y) is a point on the parabola, then PS = distance of P from the directrix.
⇒ \(\sqrt{(x-0)^2+(y+3)^2}\) = (y -3)2
⇒ x2 + y2 + 6y + 9 = y2 – 6y + 9
⇒ x2 = -12y.
Equation of the parabola is x2 = -12y.
Question 3.
If the parabola y2 = 4ax passes through the point (3, 2) then the length of its latusrectum is
1) \(\frac{2}{3}\)
2) \(\frac{4}{3}\)
3) \(\frac{1}{3}\)
4) 4
Answer:
2) \(\frac{4}{3}\)
Explanation:
Parabola y2 = 4ax passes through the point (3, 2)
⇒ (2)2 = 4a(3)
⇒ 4a = \(\frac{4}{3}\)
Length of latusrectum = 4a = \(\frac{4}{3}\)
Question 4.
If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0 then its equation is
1) y2 = 8(x + 3)
2) x2 = 8(y + 3)
3) y2 = -8(x + 3)
4) y2 = 8(x + 5)
Answer:
1) y2 = 8(x + 3)
Explanation:
Given equation of the directrix is x = – 5 which is parallel to Y-axis.
∴ Principal axis of the parabola is parallel to X-axis. Vertex = (-3, 0).
Equation of the parabola is (y – 0)2 = 4a(x + 3) ⇒ y2 = 4a(x + 3)
We know the equation of the directrix is x = h – a ⇒ x = – 3 – a, but given as x = – 5.
∴ – 5 = -3 – a ⇒ a = 5 – 3 = 2
Equation of the parabola is y2 = 8(x + 3).
Ellipse
Question 1.
The eccentricity of an ellipse, when its centre at the origin is \(\frac{1}{2}\). If one of the directrices is x = 4, then the equation of the ellipse is ( 4 )
1) 3x2 + 4y2 = 1
2) 4x2 + 3y2 = 1
3) 4x2 + 3y2 = 12
4) 3x2 + 4y2 = 12
Answer:
4) 3x2 + 4y2 = 12
Explanation:
Centre = (0, 0), Eccentricity e = \(\frac{1}{2}\) and directrix is x = 4
⇒ \(\frac{a}{e}\) = 4
⇒ 2a = 4
⇒ a = 2
b2 = a2(1 – e2) = 4(1 – \(\frac{1}{4}\)) = 3 ;
Equation of the ellipse is \(\frac{x^2}{4}+\frac{y^2}{3}\) = 1
⇒ 3x2 + 4y2 = 12
Question 2.
The equation of the ellipse whose focus is at (4, 0) and whose eccentricity \(\frac{4}{5}\) is
1) \(\frac{x^2}{3^2}+\frac{y^2}{5^2}\) = 1
2) \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1
3) \(\frac{x^2}{5^2}+\frac{y^2}{4^2}\) = 1
4) \(\frac{x^2}{4^2}+\frac{y^2}{5^2}\) = 1
Answer:
2) \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1
Explanation:
e = \(\frac{4}{5}\), (ae, 0) = (4, 0)
⇒ ae = 4
⇒ a = 5 ; b2 = 25(1 – \(\frac{16}{25}\)) = 9
∴ Equation of the ellipse is \(\frac{x^2}{25}+\frac{y^2}{9}\) = 1
⇒ \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1
Question 3.
The ellipse x2 + 4y2 = 4 is inscribed is a rectangular (rectangle) aligned with the co-ordinate axes which in turn is inscribed in another. The equation of the ellipse is
1) x2 + 12y2 = 16
2) 4x2 – 8y2 = 48
3) 4x2 + 64y2 = 48
4) x2 + 16y2 = 16
Answer:
1) x2 + 12y2 = 16
Explanation:
The given ellipse is x2 + 4y2 = 4 ⇒ \(\frac{x^2}{4}+\frac{y^2}{1}\) = 1.
The vertex of the rectangle (2, 1) is in I quadrant. Again, the ellipse circumscribing the rectangle passes through the point (4, 0). So its equation is \(\frac{x^2}{16}+\frac{y^2}{b^2}\) = 1
A(2, 1) lies on the above ellipse, \(\frac{4}{16}+\frac{1}{b^2}\) = 1
⇒ \(\frac{1}{\mathrm{~b}^2}\) = 1 – \(\frac{1}{4}\)
⇒ b2 = \(\frac{4}{3}\)
Thus the equation of the desired ellipse is \(\frac{x^2}{16}+\frac{3}{4}\)y2 = 1
x2 + 12y2 = 16
Question 4.
If focus of an ellipse is at the origin the directrix is the line x = 4 and the eccentricity is \(\frac{1}{2}\). Then the length of the semi-major axis is
1) \(\frac{2}{3}\)
2) \(\frac{4}{3}\)
3) \(\frac{5}{3}\)
4) \(\frac{8}{3}\)
Answer:
4) \(\frac{8}{3}\)
Explanation:
Given S = (0, 0), e = \(\frac{1}{2}\), equation to directrix is x – 4 = 0.
We have CZ = \(\frac{a}{e}\)
⇒ CS + SZ = \(\frac{a}{e}\)
⇒ ae + SZ = \(\frac{a}{e}\)
⇒ SZ = \(\frac{a}{e}\) – ae.
Distance from focus to directrix = a(\(\frac{1}{e}\) – e)
|0 – 4| = a(2 – \(\frac{1}{2}\)) ⇒ a\(\left(\frac{3}{2}\right)\) = 4
⇒ a = \(\frac{8}{3}\).
∴ Length of the semi major axis is a = \(\frac{8}{3}\).
Question 5.
The major axes of an ellipse is three times minor axis, then the eccentricity is
1) \(\frac{2 \sqrt{2}}{3}\)
2) \(\frac{2}{3}\)
3) \(\frac{\sqrt{2}}{3}\)
4) \(\frac{1}{3}\)
Answer:
1) \(\frac{2 \sqrt{2}}{3}\)
Explanation:
2a = 3 . 2b ⇒ a = 3b
∴ e2 = \(\frac{a^2-b^2}{a^2}=\frac{8 b^2}{9 b^2}=\frac{8}{9}\)
e = \(\frac{2 \sqrt{2}}{3}\)
Question 6.
The eccentricity of ellipse x2 + 3y2 = 6 is
1) \(\frac{2}{3}\)
2) \(\sqrt{\frac{2}{3}}\)
3) \(\frac{\sqrt{3}}{2}\)
4) \(\frac{2}{\sqrt{3}}\)
Answer:
2) \(\sqrt{\frac{2}{3}}\)
Explanation:
Given ellipse is x2 + 3y2 = 6
⇒ \(\frac{x^2}{6}+\frac{y^2}{2}\) = 1 (a > b).
Eccentricity of the ellipse = \(\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{6-2}{6}}=\sqrt{\frac{4}{6}}=\sqrt{\frac{2}{3}}\)
Hyperbola
Question 1.
The equation of the conic with focus at (1, – 1), directrix along x – y + 1 = O and with eccentricity √2 is
1) x2 – y2 = 1
2) xy = 1
3) 2xy – 4x + 4y + 1 = 0
4) 2xy + 4x – 4y + 1 = 0
Answer:
3) 2xy – 4x + 4y + 1 = 0
Explanation:
Focus S = (1, – 1), eccentricity e = √2, directrix along x – y + 1 = 0.
Let (x, y) be a point on the hyperbola ⇒ SP = e PM
⇒ SP2 = e2 PM2
⇒ (x – 1)2 + (y + 1)2 = 2\(\frac{1}{2}\)
⇒ x2 + y2 – 2x + 2y + 2 = x2 + y2 + 1 – 2xy – 2y + 2x
⇒ 2xy – 4x + 4y + 1 = 0, which is the equation of the conic.
Question 2.
If one of the foci of the hyperbola is origin the corresponding directrix is 3x + 4y + 1 = 0 and the eccentricity of the hyperbola is √5, then equation of the hyperbola is
1) 4x2 + 11y2 + 24xy + 6x + 8y + 1 = 0
2) 8x2 + 9y2 + 24xy + 6x + 6y + 1 = 0
3) 8x2 + 9y2 + 24xy + 6x + 8y + 1 = 0
4) 8x2 + 9y2 + 24xy + 6x + 8y + 1 = 0
Answer:
1) 4x2 + 11y2 + 24xy + 6x + 8y + 1 = 0
Explanation:
Let P(x1, y1) be a point in the hyperbola. The distance from P to origin is \(\sqrt{x_1^2+y_1^2}\).
The distance from P to the d.irectrix is \(\left|\frac{3 x_1+4 y_1+1}{5}\right|\)
∴ \(\sqrt{x_1^2+y_1^2}=\sqrt{5}\left|\frac{3 x_1+4 y_1+1}{5}\right|\)
⇒ 5(x12 + y12) = (3x1 + 4y1 + 1)
⇒ 5x12 + 5y12 = 9x12 + 16y1 + 1 + 24x1y1 + 6x1 + 8y1
⇒ 4x12 + 24x1y1 + 11y12 + 6x1 + 8y1 + 1—O
∴ The equation of the hyperbola is 4x2 + 11y2 + 24xy + 6x + 8y + 1 – 0
Question 3.
The vertices of the hyperbola are (2, 0), (-2, 0) and the foci are (3, 0), (-3, 0) the equation of the hyperbola is
1) \(\frac{x^2}{5}-\frac{y^2}{4}\) = 1
2) \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1
3) \(\frac{x^2}{5}-\frac{y^2}{2}\) = 1
4) \(\frac{x^2}{2}-\frac{y^2}{5}\) = 1
Answer:
2) \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1
Explanation:
Vertices of the hyperbola = (±2, 0) ⇒ a = 3
Foci of the hyperbola = (±3, 0) ⇒ ae – 3 ⇒ 2e = 3 ⇒ e = \(\frac{3}{2}\)
b2 – a2(e2 – 1)
⇒ b2 – 4(\(\frac{9}{4}\) – 1)
⇒ b2 = 5.
Equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1
Question 4.
The foci of the hyperbola 9y2 – 4x2 = 36 are the points ( 1 )
1) (0, ±√13)
2) (±3, 0)
3) (0, ±3)
4) (±13, 0)
Answer:
1) (0, ±√3)
Explanation:
Given hyperbola is 9y2 – 4x2 = 36
Eccentricity, e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{9+4}{4}}=\frac{\sqrt{13}}{2}\)
Foci of the hyperbola = (0, ±be) = (0, ±2 × \(\frac{\sqrt{13}}{2}\)) = (0, ±\(\sqrt{13}\))
Question 5.
The distance between the foci of the hyperbola x2 – 3y2 – 4x – 6y – 11 = o is
1) 4
2) 6
3) 8
4) 10
Answer:
3) 8
Explanation:
x2 – 3y2 – 4x – 6y – 11 = o
⇒ (x2 – 4x + 4) – 3(y2 + 2y + 1) = 12
⇒ (x – 2)2 – 3(y + 1)2 = 12
⇒ \(\frac{(x-2)^2}{12}-\frac{(y+1)^2}{4}\) = 1
⇒ a2 = 12, b2 = 4
e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\frac{\sqrt{12+4}}{\sqrt{12}}=\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}}\)
Distance between foci = 2ae = 2\(\sqrt{12}\left(\frac{2}{\sqrt{3}}\right)\) = 8.
Question 6.
The equation \(\frac{1}{2}\) = 1 represents a hyperbola if c ( 1 )
1) lies between 5 and 9
2) does not lie between 5 and 9
3) lies between 0 and 5
4) lies between 0 and 9
Answer:
1) lies between 5 and 9
Explanation:
\(\frac{x^2}{9-c}+\frac{y^2}{5-c}\) = 1 represents a hyperbola if
9 – c > 0 ⇒ 9 > c; 5 – c < 0 ⇒ 5 < c
∴ 5 < c < 9.
Question 7.
If e and e’ are the eccentricities of the ellipse 5x2 + 9y2 = 45 and the hyperbola 5x2 – 4y2 = 45 respectively then ee’
1) 9
2) 5
3) 4
4) 1
Answer:
4) 1
Explanation:
5x2 + 9y2 = 45
⇒ \(\frac{x^2}{9}+\frac{y^2}{5}\) = 1
⇒ e = \(\frac{\sqrt{9-5}}{3}=\frac{2}{3}\)
5x2 – 4y2 = 45
⇒ \(\frac{x^2}{9}-\frac{y^2}{\frac{45}{4}}\) = 1
e’ = \(\frac{\sqrt{9+\frac{45}{4}}}{3}=\frac{9}{6}=\frac{3}{2}\)
∴ ee’ = \(\frac{2}{3} \times \frac{3}{2}\) = 1
Question 8.
If the foci of the ellipse \(\frac{1}{2}\) = 1 and the hyperbola \(\frac{1}{2}\) = 1 coincide then b2 =
1) 4
2) 5
3) 8
4) 9
Answer:
2) 5
Explanation:
Eccentricity of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1 is \(\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{25-16}{25}}=\frac{3}{5}\). Foci are (3, 0), (-3, 0).
Eccentricity of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{b^2}\) = 1 is . Foci are (\(\sqrt{\frac{4+b^2}{4}}\), 0). (-\(\sqrt{\frac{4+b^2}{4}}\), 0).
Given that S = \(\sqrt{4+b^2}\)
⇒ 9 = 4 + b2
⇒ b2 = 5.
Question 9.
For the hyperbola \(\frac{x^2}{\cos ^2 \alpha}+\frac{y^2}{\sin ^2 \alpha}\) = 1 which of the following remains constant when a varies?
1) Eccentricity
2) Directrix
3) Abscissae of vertices
4) Abscissae of foci
Answer:
4) Abscissae of foci
Explanation:
\(\frac{x^2}{\cos ^2 \alpha}+\frac{y^2}{\sin ^2 \alpha}\) = 1. We have sin2α = cos2α (e2 – 1)
tan2α = e2 – 1
⇒ e2 = sec2α
⇒ e = sec α ……….(1)
Equation of the directrix is x = a/e.
⇒ x = \(\frac{\cos \alpha}{\sec \alpha}\)
⇒ x = cos2α ……….(2)
Vertices = (± cos α, 0) …………. (3)
Foci = (±ae, 0) = (± cos α sec α, 0) = (±1, 0) …………(4)
Abscissae of foci are constant.