Equation of Ellipse MCQ Quiz in मल्याळम - Objective Question with Answer for Equation of Ellipse - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:         

Concept:

The standard Equation of Ellipse is \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Major Axis: x-axis

c² = a² - b² 

Coordinate of foci (± c, 0)

Vertices (±a, 0)

Major axis = 2a

Minor Axis = 2b

Eccentricity = e = \(\frac{c}{a}\)

Latus Rectum = \(\frac{2b^2}{a}\)

Major Axis: y-axis

c2 = b2 - a2 

Coordinate of foci (0, ± c)

Vertices (0, ±b)

Major axis = 2b

Minor Axis = 2a

Eccentricity = e = \(\frac{c}{b}\)

Latus Rectum = \(\frac{2a^2}{b}\)

Calculation:

Given:

Equation of the ellipse is \(\rm \frac{x^2}{100}+\frac{y^2}{400}=1\)

Here, a² = 100 and b² = 400

a = 10 and b = 20

Here, y-axis is the major axis.

Statement I:  Vertices of the ellipse are (0, ± 20)

Vertices (0, ±b) = (0, ± 20)

Statement I is correct.

Statement II: Foci of the ellipse are (0. ±10√3)

c2 = b2 - a2 = 400 - 100 = 300

c = √300 = 10√3

Coordinate of foci (0, ± c) = (0, ±10√3)

Statement II is correct.

Statement III:  Length of major axis is 40.

Major axis = 2b = 40

Statement III is correct.

Statement IV: Eccentricity of the ellipse is \(\frac{√3}{2}\)

Eccentricity = e = \(\frac{c}{b}\)

e = \(\frac{10√3}{20}\) = \(\frac{√3}{2}\)

Statement IV is correct.

∴ All the statements are correct.

Calculation:

PS + PS' = 2 × 3 √2 

⇒ b² = a² (1 - e²) ⇒ 9 = 18( 1 - e²)

⇒ e = \(\frac{1}{\sqrt2}\)

Diretrix x = \(\frac{a}{e} = \frac{3\sqrt2}{\frac{1}{\sqrt2}{}} = 6\)

⇒ PS.PS' = \(PS.PS' = | \frac{1}{\sqrt{2}} (3\sqrt{2} \cos \theta - 6)|\)

 = \(\frac{1}{2}| 18 cos^2\theta - 36|\)

⇒ (PS. PS')_max = 18 and (PS .PS')_min = 9

Sum = 27

Hence, the correct answer is Option 4.

Concept:

Equation of ellipse: \(\rm\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)​

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

Where, vertices = (± a, 0) and focus = (± ae, 0)

Calculation:

Here, vertices of ellipse (± 5, 0) and foci (±4, 0)

So, a = ±5 ⇒ \(a^2=25\) and

ae = 4 ⇒ e = 4/5

Now, 4/5 = \(\rm\sqrt{1-\frac{b^2 }{5^2}}\)

\(⇒ \rm\frac{16}{25}=\rm\frac{25-b^2}{25}\\⇒ 16=25-b^2 \\⇒ b^2=9 \)

∴ Equation of ellipse = \(\rm \frac {x^2}{25} + \frac {y^2}{9} = 1\)

Hence, option (1) is correct. 

CONCEPT:

The properties of a vertical ellipse \(\frac{{{x^2}}}{b^2} + \frac{{{y^2}}}{a^2} = 1\) where 0 < b < a are as follows:

• Centre of ellipse is (0, 0)
• Vertices of ellipse are: (0, - a) and (0, a)
• Foci of ellipse are: (0, - ae) and (0, ae)
• Length of major axis is 2a
• Length of minor axis is 2b
• Eccentricity of ellipse is given by: \(e = \frac{{\sqrt {{a^2} - {b^2}} }}{a}\) 

CALCULATION:

Given: Equation of ellipse is \(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1\)

As we can see that the given ellipse is a vertical ellipse.

By comparing the given equation of ellipse with \(\frac{{{x^2}}}{b^2} + \frac{{{y^2}}}{a^2} = 1\) where 0 < b < a we get,

⇒ a = 6, b = 2

As we know that, eccentricity of ellipse is given by: \(e = \frac{{\sqrt {{a^2} - {b^2}} }}{a}\)

⇒ \(e = \frac{{\sqrt {{6^2} - {2^2}} }}{6} = \frac{2\sqrt 2}{3}\)

⇒ ae = 4√2

As we know that, foci of a vertical ellipse are: (0, - ae) and (0, ae)

So, the foci of the given ellipse are: (0, - 4√2) and (0, 4√2)

Hence, option D is the correct answer.

Concept:

The distance between the centre and the focus of an ellipse is c = ae

The equation of an ellipse with the length of the major axis 2a and the minor axis 2b is given by: \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).

Calculation:

Length of the minor axis = 2b = 8.

⇒ b = 4.

Also, c = distance between the centre and the focus = ae = 2.

c2 = a2e2 = a2 - b2

∴ 22 = a2 - 42

⇒ a² = 20

Equation of the ellipse = \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).

⇒ \(\rm \frac{x^2}{20}+\frac{y^2}{16}=1\).

Concept:

Eccentricity: the ratio of the distance between a point on the ellipse and its focus to the distance between the point and the directrix is called its eccentricity.

Calculation:

Let P(x, y) be a point on the ellipse, F(1, 0) be its focus and D = x + y + 1 = 0 be its directrix.

PF = e × PD

⇒ PF2 = e² × PD2

⇒ (x - 1)2 + (y - 0)2 = \(\rm \frac12 \times\left(\frac{x+y+1}{\sqrt{1^2+1^2}}\right)^2\)

⇒ 4(x² - 2x + 1 + y2) = x2 + y2 + 1 + 2xy + 2x + 2y

⇒ 3x2 + 3y2 - 2xy - 10x - 2y + 3 = 0

Concept Used:

The standard equation of an ellipse with major axis along the x-axis is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where a > b.

The foci of the ellipse are at (±c, 0), where c² = a² - b².

Calculation

Given: The equation of the ellipse is \(\frac{x^2}{49} + \frac{y^2}{24} = 1\).⇒ a² = 49 and b² = 24

⇒ a = 7 and b = \(\sqrt{24}\)

⇒ c² = a² - b² = 49 - 24 = 25

⇒ c = \(\sqrt{25}\) = 5

⇒ The foci are at (±5, 0).

∴ The foci of the ellipse are (5, 0) and (-5, 0).

The correct option is 4) (5, 0) and (-5, 0).

Calculation

\(eq1:\quad (x-1)^2+y^2=1\)

Diameter of this circle =2

\(eq2:\quad x^2+(y-2)^2=4\)

Diameter of this circle =4

Length of semi-minor axis =2 [along X-axis]

Length of semi major axis =4 [along Y-axis]

Required ellipse \(eq3:\quad \dfrac{x^2}{2^2}+\dfrac{y^2}{4^2}=1\Rightarrow 4x^2+y^2=16\)

Hence option 3 is correct

(here \( a < b \))

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

\( \therefore b^{2} = 8 \)

\( \therefore \) equation of ellipse \( \dfrac{x^{2}}{4} + \dfrac{y^{2}}{8} = 1 \).