Eccentricity of a conic MCQ Quiz - Objective Question with Answer for Eccentricity of a conic - Download Free PDF

Calculation:

Given any point on the ellipse is 3sinα, 5cosα. In the standard parametric form of an ellipse,

$x=a\sin \alpha ,y=b\cos \alpha$

we identify

$a=3,b=5$

Since (b > a), the semi-major axis is (b = 5) and the semi-minor axis is (a = 3). The eccentricity e of an ellipse is

$e=\sqrt{1-\frac{(\text{semi-minor}{)}^2}{(\text{semi-major}{)}^2}}=\sqrt{1-\frac{a^2}{b^2}}$

Substitute (a = 3) and (b = 5):

$e=\sqrt{1-\frac{3^2}{5^2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Hence, the correct answer is Option 2.

Concept:

Equation of ellipse: $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

Eccentricity = $\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

Calculation:

Here, 4x2 + 9y2 = 144

$$\begin{gathered} \Rightarrow \frac{4{\mathrm{x}}^2}{144}+\frac{9{\mathrm{y}}^2}{144}=1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{\frac{144}{4}}+\frac{{\mathrm{y}}^2}{\frac{144}{9}}=1 \end{gathered}$$

So, a² = 144/4 and b² = 144/9

∴Eccentricity

 $$\begin{gathered} \sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}} \\ =\sqrt{1-\frac{144}{9}\times \frac{4}{144}} \\ =\sqrt{1-\frac{4}{9}} \\ =\sqrt{\frac{5}{9}} \\ =\frac{\sqrt{5}}{3} \end{gathered}$$

Hence, option (1) is correct. 

Calculation 

Product of focal distances = (a + ex₁) (a – ex₁)

⇒ ${\mathrm{a}}^2-{\mathrm{e}}^2{\mathrm{x}}_1^2={\mathrm{a}}^2-{\mathrm{e}}^2(3)$

⇒ ${\mathrm{a}}^2-3{\mathrm{e}}^2=\frac{7}{4}\Rightarrow {\mathrm{a}}^2=\frac{7}{4}+3{\mathrm{e}}^2$

⇒ 4a² = 7 + 12e²

& $(\sqrt{3},\frac{1}{2})\text{ lines on }\frac{x^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

∴ $\frac{3}{{\mathrm{a}}^2}+\frac{1}{4{\mathrm{b}}^2}=1$

⇒ $\frac{3}{{\mathrm{a}}^2}+\frac{1}{4\left({\mathrm{a}}^2\right)(1-{\mathrm{e}}^2)}=1$

⇒ 12(1 – e²) + 1 = 4a² (1 – e²)

⇒ 13 – 12 e² = (7 + 12e²) (1 – e²) 

⇒ 13 – 12 e² = 7 – 7e² + 12e² – 12e⁴ 

⇒ 12 e⁴ – 17e² + 6 = 0

∴ ${\mathrm{e}}^2=\frac{17\pm \sqrt{289-288}}{24}=\frac{17\pm 1}{24}=\frac{3}{4}\mathrm{\&}\frac{2}{3}$

∴ $\mathrm{e}=\frac{\sqrt{3}}{2}\mathrm{\&}\sqrt{\frac{2}{3}}$

∴ difference = $\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2\sqrt{2}}{2\sqrt{3}}$

Hence option 3 is correct

Calculation 

Given $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\mathbf{1}$

At (9, 5)

$\frac{81}{a^2}+\frac{25}{b^2}=1$ ......(1)

At (12,4)

$\frac{144}{a^2}+\frac{16}{b^2}=1$ .......(2)

From eq. (2) - eq. (1):

$\frac{63}{a^2}-\frac{9}{b^2}=0$

⇒ $\frac{b^2}{a^2}=\frac{1}{7}$

⇒ $\frac{b}{a}=\frac{1}{\sqrt{7}}$

e = $\sqrt{1-\frac{b^2}{a^2}}$ = $\sqrt{1-\frac{1}{7}}$ = $\sqrt{\frac{6}{7}}$

Hence option 4 is correct

Explanation:

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

Case 1: if a = b= 0, 

 2hxy + 2gx + 2fy + c = 0

It is a degenerate conic or a pair of straight lines.

Case 2: a = b $\ne$ 0 and h=0

ax2 +  by2 + 2gx + 2fy + c = 0

the equation reduces to the form of a circle's equation in standard form 

Case 3: a = b $\ne$ 0 and h $\ne$ 0

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

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ax2 + 2hxy + by2 = 0

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$Hence\; option\; 2\; is\; correct$

Concept:

Standard equation of an hyperbola : $\frac{{\mathbf{x}}^2}{{\mathbf{a}}^2}-\frac{{\mathbf{y}}^2}{{\mathbf{b}}^2}=1$ 

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$ ⇔ a2e2 = a2 + b2
• Length of Latus rectum = $\frac{2{\mathrm{b}}^2}{\mathrm{a}}$

Calculation:

Given: $\frac{{\mathrm{x}}^2}{100}-\frac{{\mathrm{y}}^2}{75}=1$

Compare with the standard equation of a hyperbola: $\frac{{\mathbf{x}}^2}{{\mathbf{a}}^2}-\frac{{\mathbf{y}}^2}{{\mathbf{b}}^2}=1$

So, a2 = 100 and b2 = 75

Now, Eccentricity (e) = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$ 

= $\sqrt{1+\frac{75}{100}}$

= $\sqrt{1+\frac{3}{4}}$

= $\sqrt{\frac{7}{4}}$

Concept:

The standard equation of the ellipse is $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$.

• For any ellipse the eccentricity is always less than 1.
• For any ellipse the following relation holds for:${\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)$

Calculation:

The given equation of the ellipse is 2x² + 3y² = 6.

Divide throughout by 6 to get the equation in the standard form.

$\frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{2}=1$

Therefore, a² = 3 and b² = 2.

Now, for any ellipse using the relation we can write:

$2=3\left(1-{\mathrm{e}}^2\right)$

e² = $\frac{1}{3}$

e = $\frac{1}{\sqrt{3}}$

Therefore, the eccentricity of the given ellipse is $\frac{1}{\sqrt{3}}$. 

Concept: 

• Equation of Hyperbola , $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ , Eccentricity, e = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$
• Equation of Hyperbola , $-\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ , Eccentricity, e = $\sqrt{1+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}$ 

​Calculation: 

Equation of give Hyperbola is  $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{16}=1$, 

On comparing with standard equation , a = 2 and  b = 4 

We know that Eccentricity, e = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$ 

⇒ e = $\sqrt{1+\frac{4^2}{2^2}}$ 

⇒ e = $\sqrt{5}$ . 

The correct option is 1. 

CONCEPT:

The following are the properties of a horizontal ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where 0 < b < a:

• Its centre is (0, 0)
• Its vertices are (- a, 0) and (a, 0)
• Its foci are (- ae, 0) and (ae, 0)
• Length of the major axis is 2a
• Length of the minor axis is 2b
• Equation of major axis is y = 0
• Equation of minor axis is x = 0
• Length of the latus rectum is given by $\frac{2b^2}{a}$
• Eccentricity is given by $e=\frac{\sqrt{a^2-b^2}}{a}$

CALCULATION:

Here, we have to find the eccentricity of the ellipse whose vertices are at (± 6, 0) and foci at (± 4, 0)

As we know that, for a horizontal ellipse i.e $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we have foci at (± ae, 0) and vertices at (± a, 0)

So, by comparing the given vertices at (± 6, 0) and foci at (± 4, 0) with vertices at (± a, 0) and  foci at (± ae, 0) respectively we get

⇒ a = 6 and ae = 4

⇒ e = 4/6 = 2/3

Hence, option A is the correct answer.

Concept:

The standard equation of the ellipse is $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$.

• For any ellipse, the eccentricity is always less than 1.
• For any ellipse the following relation holds for: ${\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)$ 

Calculation:

The given equation of the ellipse is 4x² + 8y² = 24.

Now, Convert equation in the standard form.

$\frac{{\mathrm{x}}^2}{6}+\frac{{\mathrm{y}}^2}{3}=1$

Therefore, a² = 6 and b² = 3.

Now, for any ellipse using the relation we can write:

$3=6\left(1-{\mathrm{e}}^2\right)$

e² = 1/2

e = $\frac{1}{\sqrt{2}}$

Concept:

Ellipse:

Calculation:

Given: $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{25}=1$

Compare with the standard equation $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

So, a² = 16 and b² = 25 ⇔ a = 4 and b = 5 (b > a)

So, eccentricity = $\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}$

= $\sqrt{1-\frac{16}{25}}$

= 3/5

Concept:

Standard equation of Ellipse, $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$    

Eccentricity, e = $\sqrt{1-\left(\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)}$ ,  when a ≥ b 

Calculation:

x² + 2x + 2y² + 4y - 13 = 0

This can be written as:

x2 + 2x + 1 + 2 (y2 + 2y) - 13 - 1= 0

x2 + 2x + 1 + 2 (y2 + 2y + 1 - 1) - 13 - 1= 0

x2 + 2x + 1 + 2 (y2 + 2y + 1 - 1) - 13 - 1 - 2 = 0

x2 + 2x + 1 + 2 (y2 + 2y + 1) = 16

(x + 1)² + 2(y + 1)² = 16

$\frac{(\mathrm{x}+1{)}^2}{16}+\frac{(\mathrm{y}+1{)}^2}{8}$ = 1 

On comparing with standard eqn. a = 4 and b = 2√2

We know that , eccentricity , e = $\sqrt{1-\left(\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)}$   

⇒ e = $\sqrt{1-\left(\frac{8}{16}\right)}$  

⇒ e = $\frac{1}{\sqrt{2}}$ 

Explanation:

Eccentricity: Locus of a point in which the distances to the fixed point(focus) and the fixed line(directrix) are in the constant ratio. That ratio is known as eccentricity.

It is denoted by "e"

$\mathrm{e}=\frac{\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{c}\mathrm{u}\mathrm{s}}{\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}}$

Hence from the definition, e = $\frac{12}{12}$ = 1.

Hence the value of eccentricity is equal to unity.

Concept:

A standard form of the hyperbola is, 

$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}$= 1, whose eccentricity (e) = $\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

Calculation:

Given: $\frac{{\mathrm{x}}^2}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^2\alpha }-\frac{{\mathrm{y}}^2}{{\sec }^2\alpha }=1$

a² = cosec² α  and b² = sec² α 

We know that, 

b2 = a2 (e2 - 1)

sec2 α  = cosec2 α (e2 - 1)

$\frac{\mathrm{s}\mathrm{e}{\mathrm{c}}^2\alpha }{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^2\alpha }=({\mathrm{e}}^2-1)$

${\tan }^2\alpha +1={\mathrm{e}}^2$

${\sec }^2\alpha ={\mathrm{e}}^2$

$\therefore \mathrm{e}=\sec \alpha$

I. The equation of a hyperbola with the centre at origin and foci on x – axis is given by:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the eccentricity $e=\sqrt{1+\frac{b^2}{a^2}}$

II. The equation of a hyperbola with the centre at origin and foci on y – axis is given by:

$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ and the eccentricity $e=\sqrt{1+\frac{a^2}{b^2}}$

Calculation:

Given: Equation of hyperbola is 16x² – 9y² = 1

The given equation of hyperbola can be written as: $\frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1$

Here, a² = 1/16 and b² = 1/9.

As we know that, the equation of a hyperbola with the centre at origin and foci on x – axis is given by:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the eccentricity $e=\sqrt{1+\frac{b^2}{a^2}}$