Conic Sections Class 11 Notes AP Inter 1st Year Maths Chapter 10

Students can go through AP Inter 1st Year Maths Notes 10th Lesson Conic Sections will help students in revising the entire concepts quickly.

Conic Sections Class 11 Notes AP Inter 1st Year Maths 10th Lesson

→ A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.

→ The equation of a circle with centre (h, k) and the radius r is (x – h)² + (y – k)² = r².

→ A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.

→ The equation of the parabola with focus at (a, 0), a > 0, and directrix x = -a is y² = 4ax.

→ Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus, and whose endpoints lie on the parabola.

→ Length of the latus rectum of the parabola y² = 4ax is 4a.

→ The equation of a parabola when its axis is parallel to the x-axis is x = ly² + my + n.

→ An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.

→ The equation of a parabola when its axis is parallel to the y-axis is y = lx² + mx + n.

→ The equation of an ellipse with foci on the x-axis is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.

→ Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose endpoints lie on the ellipse.

→ Length of the latus rectum of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) is \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\).

→ The eccentricity of an ellipse is the ratio between the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse.

→ A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

→ The equation of a hyperbola with foci on the x-axis is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1.

→ Latus rectum of a hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose endpoints lie on the hyperbola.

→ Length of the latus rectum of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) 1 is \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\).

→ The eccentricity of a hyperbola is the ratio of the distances from the centre of the hyperbola to one of the foci and to one of the vertices of the hyperbola.

→ An equilateral hyperbola is also known as a rectangular hyperbola.

Example Problems

Question 1.
Find an equation of the circle with centre at (0, 0) and radius r.
Solution:
Here h = k = 0.
Therefore, the equation of the circle is x² + y² = r².

Question 2.
Find the equation of the circle with centre (-3, 2) and radius 4.
Solution:
Here h = -3, k = 2, and r = 4.
Therefore, the equation of the required circle is (x + 3)² + (y – 2)² = 16.

Question 3.
Find the centre and the radius of the circle x² + y² + 8x + 10y – 8 = 0
Solution:
The given equation is (x² + 8x) + (y² + 10y) = 8
Now, completing the squares within the parentheses, we get
(x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25
⇒ (x + 4)² + (y + 5)² = 49
⇒ {x – (-4)}² + {y – (-5)}² = 7²
Therefore, the given circle has centre at (-4, -5) and radius 7.

Question 4.
Find the equation of the circle that passes through the points (2, -2) and (3, 4) and whose centre lies on the line x + y = 2.
Solution:
Let the equation of the circle be (x – h)² + (y – k)² = r²
Since the circle passes through (2, -2) and (3, 4), we have
(2-h)² + (-2 – k)² = r² ………(1)
and (3 – h)² + (4 – k)² = r² ……….(2)
Also, since the centre lies on the line x + y = 2,
we have h + k = 2 ……….(3)
Solving the equations (1), (2), and (3)
We get h = 0.7, k = 1.3 and r² = 12.58
Hence, the equation of the required circle is (x – 0.7)² + (y – 1.3)² = 12.58.

Question 5.
Find the coordinates of the focus, axis, the equation of the directrix, and the latus rectum of the parabola y² = 8x.
Solution:
The given equation involves y², so the axis of symmetry is along the x-axis.
The coefficient of x is positive, so the parabola opens to the right.
Comparing with the given equation y² = 4ax, we find that a = 2.

Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = -2.
Length of the latus rectum is 4a = 4 × 2 = 8.

Question 6.
Find the equation of the parabola with focus (2, 0) and directrix x = -2.
Solution:
Since the focus (2, 0) lies on the x-axis, the x-axis itself is the axis of the parabola.
Hence, the equation of the parabola is of the form either y² = 4ax or y² = -4ax.
Since the directrix is x = -2 and the focus is (2, 0),
the parabola is to be of the form y² = 4ax with a = 2.
Hence, the required equation is y² = 4(2)x = 8x.

Question 7.
Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).
Solution:
Since the vertex is at (0, 0) and the focus is at (0, 2), which lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is of the form x² = 4ay,
thus, we have x² = 4(2)y,
i.e., x² = 8y.

Question 8.
Find the equation of the parabola that is symmetric about the y-axis, and passes through the point (2, -3).
Solution:
Since the parabola is symmetric about the y-axis and has its vertex at the origin,
The equation is of the form x² = 4ay or x² = -4ay, where the sign depends on whether the parabola opens upwards or downwards.
But the parabola passes through (2, -3), which lies in the fourth quadrant, so it must open downwards.
Thus, the equation is of the form x² = -4ay.
Since the parabola passes through (2, -3),
we have 2² = -4a(-3),
i.e., a = \(\frac {1}{3}\)
Therefore, the equation of the parabola is x² = -4(\(\frac {1}{3}\))y
i.e., 3x² = -4y.

Question 9.
Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}\) = 1.
Solution:
Since denominator of \(\frac{x^2}{25}\) is larger than the denominator of \(\frac{y^2}{9}\), the major axis is along the x-axis.
Comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, we get a = 5, b = 3
Also c = \(\sqrt{a^2-b^2}=\sqrt{25-9}\) = 4
Therefore, the coordinates of the foci are (-4, 0) and (4, 0), vertices are (-5, 0) and (5, 0).
Length of the major axis is 10 units, length of the minor axis 2b is 6 units, and the eccentricity is \(\frac {4}{5}\), and latus rectum is \(\frac{2 b^2}{a}=\frac{18}{5}\).

Question 10.
Find the coordinates of the foci, the vertices, the length of the major and minor axes, and the eccentricity of the ellipse 9x² + 4y² = 36.
Solution:
The given equation of the ellipse can be written in standard form as \(\frac{x^2}{4}+\frac{y^2}{9}\) = 1
Since the denominator of \(\frac{y^2}{9}\) is larger than the denominator of \(\frac{x^2}{4}\), the major axis is along the y-axis.
Comparing the given equation with the standard equation \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1, we have b = 2 and a = 3
Also c = \(\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}\)
and e = \(\frac{c}{a}=\frac{\sqrt{5}}{3}\)
Hence the foci are (0, √5) and (0, -√5), vertices are (0, 3) and (0, -3), length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the ellipse is \(\frac{\sqrt{5}}{3}\).

Question 11.
Find the equation of the ellipse whose vertices are (±13, 0) and foci are (±5, 0).
Solution:
Since the vertices are on the x-axis, the equation will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, where a is the semi-major axis.
Give that a = 13, c = ±5.
Therefore, from the relation c² = a² – b²,
we get 25 = 169 – b², i.e., b = 12
Hence, the equation of the ellipse is \(\frac{x^2}{169}+\frac{y^2}{144}\) = 1.

Question 12.
Find the equation of the ellipse, whose length of the major axis is 20 and whose foci are (0, ±5).
Solution:
Since the foci are on the y-axis, the major axis is along the y-axis.
So, the equation of the ellipse is of the form \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1
Given that a = semi-major axis = \(\frac {20}{2}\) = 10
and the relation c² = a² – b² gives 5² = 10² – b² i.e., b² = 75
Therefore, the equation of the ellipse is \(\frac{x^2}{75}+\frac{y^2}{100}\) = 1.

Question 13.
Find the equation of the ellipse, with major axis along the x-axis passing through the points (4, 3) and (-1, 4).
Solution:
The standard form of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.
Since the points (4, 3) and (-1, 4) lie on the ellipse,
we have \(\frac{16}{a^2}+\frac{9}{b^2}\) = 1 ……….(1)
and \(\frac{1}{a^2}+\frac{16}{b^2}\) = 1 ……….(2)
Solving equations (1) and (2), we find that
a² = \(\frac {247}{7}\) and b² = \(\frac {247}{15}\)
Hence the required equation is \(\frac{x^2}{\left(\frac{247}{7}\right)}+\frac{y^2}{\left(\frac{247}{15}\right)}=1\)
i.e., 7x² + 15y² = 247.

Question 14.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbolas:
(i) \(\frac{x^2}{9}-\frac{y^2}{16}\) = 1
(ii) y² – 16x² = 16
Solution:
(i) Comparing the equation \(\frac{x^2}{9}-\frac{y^2}{16}\) = 1 with the standard equation \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Here, a = 3, b = 4 and c = \(\sqrt{a^2+b^2}=\sqrt{9+16}\) = 5
Therefore, the coordinates of the foci are (±5, 0) and those of the vertices are (±3, 0).
Also, The eccentricity e = \(\frac{c}{a}=\frac{5}{3}\)
The latus rectum = \(\frac{2 b^2}{a}=\frac{32}{3}\)

(ii) Dividing the equation by 16 on both sides, we have \(\frac{y^2}{16}-\frac{x^2}{1}\) = 1.
Comparing the equation with the standard equation \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1,
we find that a = 4, b = 1 and c = \(\sqrt{a^2+b^2}=\sqrt{16+1}=\sqrt{17}\)
Therefore, the coordinates of the foci are (0, ±√17), and those of the vertices are (0, ±4).
Also, The eccentricity e = \(\frac{c}{a}=\frac{\sqrt{17}}{4}\)
The latus rectum = \(\frac{2 b^2}{a}=\frac{1}{2}\)

Question 15.
Find the equation of the hyperbola with foci are (0, ±3) and vertices \(\left(0, \pm \frac{\sqrt{11}}{2}\right)\).
Solution:
Since the foci are on the y-axis, the equation of the hyperbola is of the form \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1.
Since vertices are \(\left(0, \pm \frac{\sqrt{11}}{2}\right)\)
a = \(\frac{\sqrt{11}}{2}\)
Also, since foci are (0, ±3);
c = 3 and b² = c² – a² = \(\frac {25}{4}\)
Therefore, the equation of the hyperbola is \(\frac{y^2}{\left(\frac{11}{4}\right)}-\frac{x^2}{\left(\frac{25}{4}\right)}=1\),
i.e., 100y² – 44x² = 275.

Question 16.
Find the equation of the hyperbola where the foci are (0, ±12) and the length of the latus rectum is 36.
Solution:
Since foci are (0, ±12), it follows that c = 12.
Length of the latus rectum = \(\frac{2 b^2}{a}\) = 36 or b² = 18a
Therefore c² = a² + b²
gives 144 = a² + 18a
i.e., a² + 18a – 144 = 0
So a = -24, 6
Since a cannot be negative, we take a = 6 and so b² = 108.
Therefore, the equation of the required hyperbola is \(\frac{y^2}{36}-\frac{x^2}{108}\) = 1
i.e., 3y² – x² = 108.

Question 17.
The focus of a parabolic mirror, as shown in the Figure, is at a distance of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance AB.

Solution:
Since the distance from the focus to the vertex is 5 cm.
We have a = 5.
If the origin is taken at the vertex and the axis of the mirror lies along the positive x-axis,
the equation of the parabolic section is y² = 4(5)x = 20x
Note that x = 45.
Thus y² = 900
Therefore y = ±30
Hence AB = 2y
= 2 × 30
= 60 cm.

Question 18.
A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at its centre, there is a deflection of 3 cm at the centre, and the deflected beam is in the shape of a parabola. How far from the centre is the deflection of 1 cm?
Solution:
Let the vertex be at the lowest point and the axis vertical.
Let the coordinate axis be chosen as shown in the Figure.

The equation of the parabola takes the form x² = 4ay.
Since it passes through (6, \(\frac {3}{100}\)), we have
(6)² = 4a(\(\frac {3}{100}\)),
i.e., a = \(\frac{36 \times 100}{12}\) = 300 m
Let AB be the deflection of the beam, which is \(\frac {1}{100}\) m.
Coordinates of B are (2, \(\frac {2}{100}\))
Therefore, x² = 4 × 300 × \(\frac {2}{100}\) = 24
i.e., x = √24 = 2√6 metres.

Question 19.
A rod AB of length 15 cm rests between two coordinate axes in such a way that the end point. A lies on the x-axis, and endpoint B lies on the y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse.
Solution:
Let AB be the rod making an angle θ with OX as shown in the Figure, and P(x, y) the point on it such that AP = 6 cm.

Since AB = 15 cm, we have PB = 9 cm
From P, draw PQ and PR perpendicular to the y-axis and x-axis, respectively.
From ∆PBQ, cos θ = \(\frac {x}{9}\)
From ∆PRA, sin θ = \(\frac {y}{6}\)
Since cos²θ + sin²θ = 1
or \(\left(\frac{x}{9}\right)^2+\left(\frac{y}{6}\right)^2\) = 1
or \(\frac{x^2}{81}+\frac{y^2}{36}\) = 1
Thus, the locus of P is an ellipse.

Multiple Choice Questions

Circles

Question 1.
The diameters 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
(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

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 Q with PQ = 5 then c = _____
(1) -15
(2) 20
(3) 30
(4) -20
Answer:
(4) -20

Question 3.
The abscissae of two points A and B are the roots of the equation x² + 2ax – b² = 0, and their coordinates are the roots of the equation y² + 2py – q² = 0. The radius of the circle with AB as the 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) \(\sqrt{a^2+b^2+p^2+q^2}\)

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

Question 5.
A square is inscribed in the circle x² + y² – 2x + 4y + 3 = 0. Its sides are parallel to the coordinate 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

Question 6.
The equation of a circle with centre (cos α, sin α) and radius is
(1) x² + y² = 1
(2) x² + y² – 2x cos α – 2y sin α = 0
(3) x² + y² + 2x cos α + 2y sin α = 0
(4) x² + y² – 2x cos α – 1 = 0
Answer:
(2) x² + y² – 2x cos α – 2y sin α = 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

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) \(\frac {19}{20}\)

Question 9.
The perimeter of the circle 3(x² + y²) = 16 is
(1) \(\frac{4}{\sqrt{3}} \pi\)
(2) \(\frac{16}{\sqrt{3}} \pi\)
(3) \(\frac{8}{\sqrt{3}} \pi\)
(4) \(\frac{1}{\sqrt{3}} \pi\)
Answer:
(3) \(\frac{8}{\sqrt{3}} \pi\)

Question 10.
The radius of the 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

Parabola

Question 1.
Parabola has the origin as its found 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)

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

Question 3.
If the parabola y² = 4ax passes through the point (3, 2), then the length of its latus rectum is
(1) \(\frac {2}{3}\)
(2) \(\frac {4}{3}\)
(3) \(\frac {1}{3}\)
(4) 4
Answer:
(2) \(\frac {4}{3}\)

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)

Ellipse

Question 1.
The eccentricity of an ellipse, when its centre is at the origin, is \(\frac {1}{2}\). If one of the directrices is x = 4, then the equation of the ellipse is
(1) 3x² + 4y² = 1
(2) 4x² + 3y² = 1
(3) 4x² + 3y² = 12
(4) 3x² + 4y² = 12
Answer:
(4) 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) \(\frac{x^2}{5^2}+\frac{y^2}{3^2}=1\)

Question 3.
The ellipse x² + 4y² = 4 is inscribed is a rectangular (rectangle aligned with the coordinate 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

Question 4.
If the 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) \(\frac {8}{3}\)

Question 5.
The major axes of an ellipse are three times the 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) \(\frac{2 \sqrt{2}}{3}\)

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) \(\sqrt{\frac{2}{3}}\)

Hyperbola

Question 1.
The equation of the conic with focus at (1, -1), directrix along x – y + 1 = 0, 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:
(2) xy = 1

Question 2.
If one of the foci of the hyperbola is the origin, the corresponding directrix is 3x + 4y + 1 = 0, and the eccentricity of the hyperbola is √5, then the 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

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

Question 4.
The foci of the hyperbola 9y² – 4x² = 36 are the points
(1) (0, ±√3)
(2) (±3, 0)
(3) (0, ±3)
(4) (±13, 0)
Answer:
(1) (0, ±√3)

Question 5.
The distance between the foci of the hyperbola x² – 3y² – 4x – 6y – 11 = 0 is
(1) 4
(2) 6
(3) 8
(4) 10
Answer:
(3) 8

Question 6.
The equation \(\frac{x^2}{9-c}+\frac{y^2}{5-c}=1\) represents a hyperbola if c
(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

Question 7.
If e and e’ are the eccentricities of the ellipse 5x² + 9y² = 45 and the hyperbola 5x² – 9y² = 45, respectively, then ee’
(1) 9
(2) 5
(3) 4
(4) 1
Answer:
(4) 1

Question 8.
If the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}\) = -1 and the hyperbola \(\frac{x^2}{4}-\frac{y^2}{b^2}\) = 1 coincide then b² = _____
(1) 4
(2) 5
(3) 8
(4) 9
Answer:
(2) 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 α varies?
(1) Eccentricity
(2) Directrix
(3) Abscissae of vertices
(4) Abscissae of foci
Answer:
(4) Abscissae of foci