<p><strong>Let <math><mfrac><mrow><mrow><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>+</mo><mfrac><mrow><mrow><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>=</mo><mn>1</mn><mo>(</mo><mi>a</mi><mo></mo><mi>b</mi><mo>)</mo></math> be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, <math><mi>?</mi><mo>(</mo><mi>t</mi><mo>)</mo><mo>=</mo><mfrac><mrow><mrow><mn>5</mn></mrow></mrow><mrow><mrow><mn>12</mn></mrow></mrow></mfrac><mo>+</mo><mi>t</mi><mo>-</mo><msup><mrow><mrow><mi>t</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> , then <math><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> is equal to:</strong></p>

x2a2+y2b2=1(ab);2b2a=10⇒b2=5aNow, ?(t)=512+t-t2=812-t-122?(t)max=812=23=e⇒e2=1-b2a2=49⇒a2=81 (From (i) and (ii)So, a2+b2=81+45=126

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Maths Class 12th Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a b)$ be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, $? (t) = \frac{5}{12} + t - t^{2}$ , then $a^{2} + b^{2}$ is equal to:

Option 1 -

126

Option 2 -

135

Option 3 -

116

Option 4 -

145

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a b); \frac{2 b^{2}}{a} = 10 \Rightarrow b^{2} = 5 a$

Now, $? (t) = \frac{5}{12} + t - t^{2} = \frac{8}{12} - {(t - \frac{1}{2})}^{2}$
$? (t)_{m a x} = \frac{8}{12} = \frac{2}{3} = e \Rightarrow e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{4}{9}$

$\Rightarrow a^{2} = 81$ (From (i) and (ii)

So, $a^{2} + b^{2} = 81 + 45 = 126$

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Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a b)$ be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, $? (t) = \frac{5}{12} + t - t^{2}$ , then $a^{2} + b^{2}$ is equal to:

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Let x2a2+y2b2=1(ab) be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, ?(t)=512+t-t2 , then a2+b2 is equal to:

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Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a b)$ be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, $? (t) = \frac{5}{12} + t - t^{2}$ , then $a^{2} + b^{2}$ is equal to: