Conic Sections: Important Questions

Very Short Answer Questions (1 Mark Questions)

Ques. Find the equation of the Circle with centre (A, B) and touching the y-axis.

1. x² + y² + 2By + B² = 0
2. x² + y² – 2Ax – 2By + B² = 0
3. x² + y² – 2Ax + 2By + B² = 0
4. All of the above

Ans. (b) x² + y² – 2Ax – 2By + B² = 0

Here, we shall use the equation of the circle by using the general formula of the circle which is (x – h)² + (y – k)² = r²

Ques. Find the focus of parabola, 3x2 = -8y.

1. 3x – 2 = 0, x = 0
2. 3y – 8x = 0
3. 3y – 4 = 0, x = 0
4. None of the above

Ans. (c) 3y – 4 = 0, x = 0

The equation of parabola is 3x² = -8y

Now comparing it with, x² = -4ay

4a = \(\frac{8}{3}\) => a = \(\frac{2}{3}\)

So, the focus can be (0, -a) or (0, \(-\frac{2}{3}\))

Ques. Find the eccentricity of the hyperbola, 3x² – 4y² = 5

1. e = \(\sqrt{\frac{5}{2}}\)
2. e = \(\sqrt{\frac{3}{8}}\)
3. e = \(\sqrt{\frac{7}{4}}\)
4. None of the above

Ans. (c) e = \(\sqrt{\frac{7}{4}}\)

Eccentricity is given as, e = \(\sqrt{1+ \frac{b^2}{a^2}}\)

e = \(\sqrt{1+ \frac{3}{4}}\) = \(\sqrt{\frac{7}{4}}\)

Ques. Find the circle’s equation with centre (l, m) touching x-axis

1. x² + y² – 2lx – 2my + l² = 0
2. x² + y² + 2lx – 2my + l² = 0
3. x² + y² – 2lx + 2my + l² = 0
4. x² + y² + 2lx + 2my + l² = 0

Ans. (a) x² + y² – 2lx – 2my + l² = 0

The equation of the circle = (x – h)² + (y – k)² = r²

h = n, k = m; and r = n

Then the equation of the circle is given as below,

(x – n)² + (y – m)² = n²

x² – 2nx + n² + y² – 2my + m² = n²

x² + y² – 2lx – 2my + l² = 0.

Ques. Find the length of the Latus rectum of 2x2 + 4y2 = 16

1. 2 units
2. 4 units
3. 3 units
4. 8 units

Ans. (d) 8 units

Comparing with, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\),

\(\frac{x^2}{8} + \frac{y^2}{4} = 1\),

a = 2√2, b = 2

So, length of latus rectum = \(\frac{2a^2}{b}\)

= \(\frac{2 \times (2\sqrt{2})^2}{2}\)

= 8 units.

Ques. Find the latus rectum of the parabola 3y2 = 6x

1. \(\frac{1}{2}\) units
2. \(\frac{4}{3}\) units
3. \(\frac{4}{5}\) units
4. None of the above

Ans. (a) \(\frac{1}{2}\)

For parabola the equation is, y² = 4ax

And, length of the latus rectum is expressed area as = 4a

4a = \(\frac{6}{3}\) => a = \(\frac{1}{2}\)

Thus, length of latus rectum = \(\frac{1}{2}\).

Ques. The equation x² + y² – 12x + 8y -72 = 0 depicts a circle, find the centre of circle.

1. (-6, -4)
2. (6, -4)
3. (6, 4)

Ans. (b) (6, -4)

Circle equation is, x² + y² + 2gx + 2fy + c = 0

x² + y² – 12x + 8y – 72 = 0

(x² – 12x) + (y² + 8y) – 72 = 0

(x² – 12x + 36) + (y² + 8y + 16) – 36 – 16 – 72 = 0

(x – 6)² + (y + 4)² = 124

Centre = (6, -4).

Ques. Find the co-ordinates of the vertices of a2 – b2 = 1

1. A (-1, 0), B (-1, 0)
2. A (-1, 0), B (1, 0)
3. A (1, 0), B (1, 0)
4. None of the above

Ans. (b) A (-1, 0), B (1, 0)

Solution:

\(\frac{a^2}{1} - \frac{b^2}{1} = 1\)

x = 1, y = 1

(x, 0) ≡ (±1, 0)

Read More: Vertex

Ques. Consider the circle (m + 3)² + (n – 4)² = 11. Find its centre.

Ans. Comparing it with standard form of circle at (h, k) is – (x – h)² + (y – k)² = r²

Thus, centre ≡ (-3, 4).

Ques. Find the length of the Latus rectum of the equation of parabola: a² – 2a = -2b – 5.

Ans. Given: a² – 2a = -22 – 5.

Adding one on both sides, we get;

a² – 2a + 1 = -2a – 4

(a-1)² = -2(a + 2)

Compare with standard equation, (y – k)² = -4a(x – h)

4a = 2

Thus, length of the Latus rectum = 4a = 2.

Short Answer Questions (2 Marks Questions)

Ques. If the latus rectum of the ellipse is equivalent to half of the minor axis, then its eccentricity is given as

Ans. Length of the Latus rectum is = 2 \(\frac{b^2}{a^2}\)

then half minor axis = b

as it is said that half minor axis and latus rectum is equal

2 \(\frac{b^2}{a^2}\) = b

Eccentricity, e = \(\sqrt{1-(\frac{b}{a})^2}\)

e = \(\sqrt{1-(\frac{1}{2})^2}\) = \(\sqrt{1-\frac{1}{4}}\) = \(\sqrt{\frac{3}{2}}\)

Ques. Calculate:

• What is the length of the latus rectum of the ellipse 4x² + 16y² = 0?
• Evaluate the length of the latus rectum of the ellipse for the given equation 9x² + 25y² = 144?

Ans. (a) Comparing with, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

= \(\frac{x^2}{16} + \frac{y^2}{4} = 1\)

a= 4, b = 2

length of latus rectum = 2 \(\frac{b^2}{a}\)

= 2 \(\frac{2^2}{4}\) = 2.

(b) Comparing with, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

= \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)

a= 5, b = 3

length of latus rectum = 2 \(\frac{b^2}{a}\) = 2 \(\frac{3^2}{25}\) = \(\frac{18}{25}\).

Read More: Complex Numbers and Quadratic Equations

Ques. Let’s say P and P’ are foci of the ellipse 16x² + 25y² = 1600. Then the area of the triangle formed by the foci P and P’ with the point (4√4, 5) is,

Ans. Ellipse is = \(\frac{x^2}{100} + \frac{y^2}{64} = 1\)

Let A and B be the foci for point K (4√4, 5)

Now to calculate area of triangle ABK

Area of triangle = \(\frac{1}{2}\) × base × height

= \(\frac{1}{2}\) × AB × height

Height is x-coordinate of K = 4√4

AB = 2 × 10 × \(\sqrt{1- \frac{64}{100}}\) = 2 × 10 × \(\frac{6}{10}\) = 12

Thus, area of triangle = \(\frac{1}{2}\) × 4√4 × 12 = 48.

Ques. Foci of the ellipse 3x² + 4y² – 12x – 8y + 4 = 0 are given as what?

Ans. 3x² + 4y² – 12x – 8y + 4 = 0 ------- given

3(x² – 4x) + 4(y² – 2y) = -4

Hence, 3(x² – 4x + 4) + 4 (y² – 2y + 1) = -4 +12 + 4

3(x – 2)² + 4(y – 1)² = 12

= \(\frac{(x-2)^2}{4} + \frac{(y-1)^2}{3} = 1\)

a² = 4, b² =3

thus, e = \(\sqrt{1-\frac{3}{4}}\) = \(\frac{1}{2}\)

hence, foci are (x-2, y-1) = (±ae, 0) = (±1, 0)

(x, y) = (2±1, 1) ≡ (3, 1) or (1, 1)

Equations of Ellipses

Ques. Distance between the foci of the ellipse 25x² + 16y² + 100x – 64y – 236 = 0 can be

Ans. 25x² + 16y² + 100x – 64y – 236 = 0

25(x + 2)² +16(y – 2)² = 400

\(\frac{(x+2)^2}{16} + \frac{(y-2)^2}{25} = 1\)

Distance between foci is = 2 × b × e

= 2 × 5 × \(\sqrt{\frac{25-16}{25}}\) = 6.

Read More: Distance Between Two Points

Ques. The equation of the circle with centre (3, 3) and touching the line 5x – 12y -20 = 0 is

Ans. Here, r = |\(\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\)|

= | \(\frac{5(3)-12(5)-20}{\sqrt{5^2+12^2}}\) |

r = |\(\frac{-65}{13}\)| = 5.

Hence, (x – h)² + (y – k)² = r²

(x – 3)² + (y – 3)² = 2²

= x² – 6x + 9 + y² – 6y + 9 = 4

= x² + y² – 6x – 6y + 9 + 14 = 0.

Ques. Evaluate the equation of the circle with centre (-4, 5) and touching the x-axis

Ans. We know that the equation of circle = x² +y² + 2gx + 2fy + c = 0

Centre of circle = (-g, -f) ≡ (-(-4), -5)

g = 4, f = -5

hence, x² + y² + 8x – 10y + c = 0

and equation touching x axis => g² = c => c = 16

hence, required equation => x² + y² + 8x – 10y + 16 =0

Ques. Find the co-ordinates of the foci: \(\frac{x^2}{36} + \frac{y^2}{121} =1\)

Ans. Given equation of ellipse: \(\frac{x^2}{36} + \frac{y^2}{121} =1\)

\(\frac{x^2}{6^2} + \frac{y^2}{11^2} =1\)

The denominator \(\frac{y^2}{121}\) > the denominator \(\frac{x^2}{36}\)

Thus, major axis lies on y-axis and minor on x-axis.

Hence, \(\frac{x^2}{b^2}\) + \(\frac{y^2}{a^2}\) = 1

So b² = 36 or b = 6 and a² = 121 or a = 11.

So, ae = c = \(\sqrt{a^2-b^2}\) = \(\sqrt{121-36}\) = \(\sqrt{85}\).

So, co-ordinates of foci are (0, ±√85)

Ques. Find the equation of the parabola with focus ≡ (2, -2) and directrix 2x – 3y + 4 = 0.

Ans. Given: focus, F ≡ (2, -2) and directrix, D ≡ 2x – 3y + 4 = 0.

Let P (x, y) be the point on the parabola so, FP² = PD²

(x – 2)² + (y – (-2))² = \(\frac{(2x-3y +4)^2}{2^2+3^2}\)

x² + 4 – 4x + y² + 4 + 4y = \(\frac{4x^2 + 9y^2 + 16 + 12-24+16}{13}\)

13x² + 13y² – 52x + 52y = 4x² + 9y² + 12

9x² + 4y² – 52x + 52y – 12 = 0.

So, the required equation is 9x² + 4y² – 52x + 52y – 12 = 0.

Read More: Coordinate Geometry

Long Answer Questions (3 Marks Questions)

Ques. Find the eccentricity and Latus rectum of the ellipse, 25x² + 36y² = 1

Ans. Equation of ellipse is 25x² + 36y² = 1 ---- given

Now comparing it with \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

= \(\frac{x^2}{\frac{1}{25}} + \frac{y^2}{\frac{1}{36}} = 1\)

a = \(\frac{1}{5}\), b = \(\frac{1}{6}\)

Now, Eccentricity: b² = a² (1 – e²)

=> \(\frac{1}{36}\) = \(\frac{1}{25}\) (1 – e²)

=> (1 – e²) = \(\frac{25}{36}\) => 1 – \(\frac{25}{36}\) = e²

=> e² = \(\frac{9}{36}\) => e = \(\frac{3}{6}\) = \(\frac{1}{2}\).

Latus rectum = 2\(\frac{b^2}{a}\)

= \(\frac{2 \times \frac{1}{6}}{\frac{1}{5}}\) = \(\frac{5}{3}\).

Ques. Find the radius and centre of:

1. A² + B² = 25
2. (a – 5)² + (b – 3)² = 20
3. (m – \(\frac{1}{2}\))² + (n + \(\frac{1}{3}\))² = \(\frac{1}{36}\)

Ans. (a) A² + B² = 25

Centre ≡ (0, 0)

Radius = 5

(b) (a – 5)² + (b – 3)² = 20

Centre ≡ (5, 3)

Radius = √20 = 2√5

(c) (m – \(\frac{1}{2}\))² + (n + \(\frac{1}{3}\))² = \(\frac{1}{36}\)

Centre ≡ (\(\frac{1}{2}\), -\(\frac{1}{3}\))

Radius = \(\frac{1}{6}\)

Read More: Circumference of a Circle

Ques. The points given (0, \(\frac{4}{3}\)), (1, 2), and (82, 30) are vertices of?

Ans. Let’s assume triangle with points as a, b, and c

Hence, ab² = (0-1)² + (\(\frac{4}{3}\) – 2)²

= 1 + \(\frac{4}{9}\) = \(\frac{13}{9}\) = 1.444

bc² = (82 – 1)² + (30 – 2)²

= 7345

ca² = (82 – 0)² + (30 – 43)²

= 14120

From here we can see that, ab² + bc² < ca²

Hence, the triangle is an Acute-angled triangle.

Ques. Show that: length of the latus rectum of the parabola (x – 2a)2 + y2 = x2 is 4a.

Ans. Given that, (x – 2a)² + y² = x²

=> x² – 4ax + 4a² + y² = x²

=> y² = 4a (x -a)

Now compare it with Y² = 4bX

Y = y, X = (x – a), b = a

As we know length of latus rectum of parabola is Y² = 4bX is 4b

Thus, length of latus rectum of parabola given is = 4 × a = 4a.

Ques. The length of the latus rectum of parabola 36[ (a – 2)² + (b – 9)²] = (5x – 6y + 12)²

Ans. The given equation can be expressed as;

(a – 2)² + (b – 9)² = \(\left( \frac{5x – 6y + 12}{\sqrt{5^2+(-6)^2}} \right)\)

Thus, the co-ordinate of the focus is (2, 9) and equation of directrix is 5x – 6y + 12 = 0

The distance of focus from the directrix is = \(\frac{|5(2) – 6(9) + 12|}{\sqrt{5^2+(-6)^2}} = \frac{32}{\sqrt{61}}\)

So, the length of the latus rectum = 2 × \(\frac{32}{\sqrt{61}}\) = \(\frac{64}{\sqrt{61}}\).

Read More: Parabola Graph: Standard Form of Parabola Equation

Ques. Find the length of the axes: 25x² – 9y² = 25?

Ans. We compare the equation with- \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Thus, \(\frac{x^2}{9} + \frac{y^2}{25} = 1\)

Here, we can see the hyperbola is horizontal

a² = 9 and b² = 25

a = 3 and b = 5

so, length of the axes = 2a and 2b

2 × 3 = 6 and 2 × 5 = 10.

Thus, length of the axes is 6 and 10.

Ques. Find the Eccentricity and co-ordinate of vertices, and co-ordinates of the foci of the equation: 36x² – 81y² = 25?

Ans. Given: 36x² – 81y² = 25

Here, we get \(\frac{x^2}{81} + \frac{y^2}{36} = 1\)

Hence, a² = 81 or a = 9

And b² = 36 or b = 6

(i) Co-ordinate of the vertices = (±a, 0)

= (±9, 0)

(ii) Co-ordinates of the foci, c = \(\sqrt{a^2+b^2}\)

c = \(\sqrt{9^2+6^2}\) = (±√117 , 0)

(iii) So, eccentricity, = \(\frac{e}{a}\) = \(\frac{\sqrt{117}}{9}\) = \(\frac{\sqrt{13}}{3}\).

Very Long Answer Questions (5 Marks Questions)

Ques. If JKLM is a quadrilateral then tan (\(\frac{J+K}{4}\)) is?

Ans. JKLM is a quadrilateral.

As we know, sum of all angles of quadrilateral = 360°.

J + K + L + M = 360°

J + K + L + M = 2(180)°

J + K = 2π – (L + M) ----- divide by ‘4’ both sides

= \(\frac{J+K}{4}\)= \(\frac{\pi}{2}\) – \(\frac{L+M}{4}\)

Now multiplying both sides by tan we get,

tan (\(\frac{J+K}{4}\)) = tan (\(\frac{\pi}{2}\) – \(\frac{L+M}{4}\))

we know, tan (\(\frac{\pi}{2}\) – ∅) = cot ∅

therefore, tan (\(\frac{J+K}{4}\)) = cot (\(\frac{L+M}{4}\)).

Ques. If WXYZ is a cyclic quadrilateral such that, 14 tan A – 5 = 0, 5 cos B + 3 = 0, then 39(cos C + tan D) will be,

Ans. 14 tan A – 5 = 0

tan A = \(\frac{5}{14}\)

5 cos B + 3 = 0

cos B = \(\frac{-3}{5}\)

∠A + ∠ C = 180°

∠A = 180° - ∠C

tan C = -\(\frac{5}{14}\)

cos C= -\(\frac{14}{13}\) ----- (13² = 14² + 5²)

now, ∠B + ∠D = 180°

∠B = 180° - ∠D

cos ∠B = cos (180° - ∠D)

cos ∠D = -cos ∠B

cos D = \(\frac{3}{5}\)

tan D = \(\frac{4}{3}\)

Thus, 39(cos C + tan D) = 39(-\(\frac{14}{13}\)+ \(\frac{4}{3}\)) = 10.

Read More: Cyclic Quadrilateral

Ques. If PQRS is a cyclic quadrilateral, tan B – tan D = 2√3, then tan 3B will be

Ans. PQRS is a cyclic quadrilateral.

∠B + ∠D = 180°

∠D = 180° - ∠B

tan B – tan D = 2√3

tan B – tan (180° – B) = 2√3

tan B + tan B = 2√3

2 tan B = 2√3 => tan B = √3

B = tan^-1 √3 => B = tan 60°

tan 3B = tan (3 x 60°)

tan 3B = tan (180°)

tan 3B = 0.

Ques. Find out what will be the co-ordinates of the focus and the vertex, the equation of the directrix and the axis, and the latus rectum of the parabola, y² = 12x.

Ans. Given, y² = 12x

So, y² = 4ax

4a = 12

a = 3

thus, at axis y = 0

vertices (0, 0)

focus = (a, 0) = (3, 0)

equation of directrix will be is x = -a

x =-3

thus, length of latus rectum = 4a = 4 × 3 = 12 units

so, F (3, 0) , O (0, 0) 3x = 0 is 12 units.

Read More: Parabola Formula: Vertex, Focus, Directrix

Ques. Prove that the equation of the circle l² + m² + l – m = 0. Find its centre and radius.

Ans. Equation of circle is given as l² + m² + l – m = 0.

We compare the given equation with general equation of circle:

x² + y² + 2gx + 2fy + c = 0, centre ≡ (-g, -f), and radius ≡ \(\sqrt{g^2+f^2-c}\)

2g = 1, 2f = -1

g = \(\frac{1}{2}\), f = -\(\frac{1}{2}\) and c = 0

thus, centre of the circle ≡ (\(\frac{1}{2}\), -\(\frac{1}{2}\))

radius ≡ \(\sqrt{(\frac{1}{2})^2+(\frac{-1}{2})^2-0} = \frac{1}{\sqrt{2}}\). Hence proved.

Ques. Show that equation 3x² + 3y² + 3x – 3y – 1 = 0 represents a circle. Find centre and radius.

Ans. The general equation is expressed as, ax² + 2hxy + by² + 2gx + 2fy + c = 0

For circle, we have ‘a’ as ‘b’ and ‘h’ as ‘0’

And standard equation of circle is x² + y² + 2gx + 2fy +c = 0. -----(1)

Taking ‘3’ common from the given equation 3x² + 3y² + 3x – 3y – 1 = 0, we get,

3(x² + y² + x – y – \(\frac{1}{3}\))= 0. -------(2)

Hence, on comparing (2) with (1)

g = \(\frac{1}{2}\), f = -\(\frac{1}{2}\), c = -\(\frac{1}{3}\)

centre ≡ (\(\frac{1}{2}\), -\(\frac{1}{2}\))

radius ≡ \(\sqrt{g^2+f^2-c}\) = \(\sqrt{(\frac{1}{2})^2+(\frac{-1}{2})^2-\frac{1}{3}} = \sqrt{\frac{1}{6}}\).