Let x = 2t, <math> <mrow> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> be a conic. Let S be the focus and B the point on the axis of the conic such that <math> <mrow> <mi>S</mi> <mi>A</mi> <mo>&perp;</mo> <mi>B</mi> <mi>A</mi> <mo>,</mo> </mrow> </math> where A is any point on the conic. If k is the ordinate of the centroid of the <math> <mrow> <mi>Δ</mi> <mi>S</mi> <mi>A</mi> <mi>B</mi> <mo>,</mo> </mrow> </math> then <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>t</mi> <mo>→</mo> <mn>1</mn> </mrow> </munder> </mrow> </math> k is equal to:

x = 2t A ( 2 t , t 2 3 ) y = t 2 3 S (0, 3) 3 y = ( x 2 ) 2 B ( 0 , λ ) 3 k = t 2 3 + 3 + λ = 2 t 2 3 + 3 − 1 2 t 2 9 − t 2 l i m t → 1 3 k = 2 3 + 3 − 1 2 8 = 3 − 5 6 = 1 3 6

Let x = 2t, <math> <mrow> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> be a conic. Let S be the focus and B the point on the axis of the conic such that <math> <mrow> <mi>S</mi> <mi>A</mi> <mo>&perp;</mo> <mi>B</mi> <mi>A</mi> <mo>,</mo> </mrow> </math> where A is any point on the conic. If k is the ordinate of the centroid of the <math> <mrow> <mi>Δ</mi> <mi>S</mi> <mi>A</mi> <mi>B</mi> <mo>,</mo> </mrow> </math> then <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>t</mi> <mo>→</mo> <mn>1</mn> </mrow> </munder> </mrow> </math> &nbsp;k is equal to:

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Maths NCERT Exemplar Solutions Class 11th Chapter Two Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to:

Option 1 -

$\frac{17}{18}$

Option 2 -

$\frac{19}{18}$

Option 3 -

$\frac{11}{18}$

Option 4 -

$\frac{13}{18}$

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Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 2

Detailed Solution:

x = 2t $A (2 t, \frac{t^{2}}{3})$

$y = \frac{t^{2}}{3}$ S (0, 3)

$3 y = {(\frac{x}{2})}^{2}$ $B (0, λ)$

$3 k = \frac{t^{2}}{3} + 3 + λ = \frac{2 t^{2}}{3} + 3 - \frac{12 t^{2}}{9 - t^{2}}$

$\underset{t \to 1}{l i m} 3 k = \frac{2}{3} + 3 - \frac{12}{8} = 3 - \frac{5}{6} = \frac{13}{6}$

x = 2t           <math> <mrow> <mi>A</mi> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mi>t</mi> <mo>, </mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>   <math> <mrow> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>          S (0, 3) <math> <mrow> <mn>3</mn> <mi>y</mi> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>           <math> <mrow> <mi>B</mi> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>, </mo> <mi>λ</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math>   <math> <mrow> <mn>3</mn> <mi>k</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mn>3</mn> <mo>+</mo> <mi>λ</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mn>3</mn> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> <mn>2</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>9</mn> <mo>−</mo> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>   <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>t</mi> <mo>→</mo> <mn>1</mn> </mrow> </munder> <mn>3</mn> <mi>k</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mn>3</mn> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> <mn>2</mn> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mo>−</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>3</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math>

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Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to:

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Let x = 2t, y = t 2 3 be a conic. Let S be the focus and B the point on the axis of the conic such that S A ⊥ B A , where A is any point on the conic. If k is the ordinate of the centroid of the Δ S A B , then l i m t → 1 k is equal to:

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Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to: