Let the maximum area of the triangle that can be inscribed in the ellipse <math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>+</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>4</mn></mrow></mfrac><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>a</mi><mtext> </mtext><mtext> </mtext><mo>></mo><mtext> </mtext><mtext> </mtext><mn>2</mn><mo>,</mo></mrow></math> having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be <math><mrow><mn>6</mn><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow></math> . Then the eccentricity of the ellipse is

x2a2+y24=1∴Δ=12×a (1+cosθ).4sinθ∴ΔmaxΔ=63⇒a=4∴e=32

Let the maximum area of the triangle that can be inscribed in the ellipse&nbsp;<math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>+</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>4</mn></mrow></mfrac><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mi>a</mi><mtext> </mtext><mtext> </mtext><mo>&gt;</mo><mtext> </mtext><mtext> </mtext><mn>2</mn><mo>,</mo></mrow></math>&nbsp;having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be&nbsp;<math><mrow><mn>6</mn><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow></math>&nbsp;. Then the eccentricity of the ellipse is

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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1, a > 2,$ having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $6 \sqrt[]{3}$ . Then the eccentricity of the ellipse is

Option 1 -

$\frac{\sqrt[]{3}}{2}$

Option 2 -

$\frac{1}{2}$

Option 3 -

$\frac{1}{\sqrt[]{2}}$

Option 4 -

$\frac{\sqrt[]{3}}{4}$

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1$

$∴ Δ = \frac{1}{2} \times a (1 + c o s θ) . 4 s i n θ$

$∴ Δ_{m a x}^{Δ} = 6 \sqrt[]{3} \Rightarrow a = 4$

$∴ e = \frac{\sqrt[]{3}}{2}$

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$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

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...more

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

Also $1 - \frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{144}{25} b^{2} + \frac{256}{25} a^{2} = a^{2} b^{2}$

$\frac{144}{25} + \frac{256}{25} \times \frac{1}{2} = a^{2}$

$\frac{(128 + 144)}{25} = a^{2} \Rightarrow \frac{272}{25} = a^{2}$

-> $b^{2} = \frac{2 \times 272}{25}$

Latus rectum = $\frac{2 b^{2}}{a}$

(Latus rectum)²

$= \frac{4 b^{4}}{a^{2}} = 4 (\frac{b^{2}}{a^{2}}) b^{2} = \frac{8 \times 272 \times 2}{25} = \frac{4352}{25}$

less

The locus of the point of intersection of the lines $(\sqrt[]{3}) k x + k y - 4 \sqrt[]{3} = 0 a n d \sqrt[]{3} x - y - 4 (\sqrt[]{3}) k$ = 0 is a conic, whose eccentricity is…………………

Vishal Baghel

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K = $\frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}$ $\frac{}{}$

$∴ \frac{4 \sqrt[]{3}}{\sqrt[]{3} x + y} = \frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}$

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48 = 16 (e2 – 1)

$e = \sqrt[]{4} = 2$

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Let the maximum area of the triangle that can be inscribed in the ellipse x2a2+y24=1, a > 2, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be 63 . Then the eccentricity of the ellipse is

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