If the tangents drawn at the points O(0, 0) and <math> <mrow> <mi>P</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>+</mo> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> on the circle x2 + y2 – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

x 2 + y 2 − 2 x − 4 y = 0 Centre (1, 2) r = 5 Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0⇒x + 2y = 0 ……… (i)Equation of PQ is x ( 1 + 5 ) + 2 y − ( | x + 1 + 5 ) − 2 ( y + 2 ) = 0 Solving (i) & (ii), Q ( 5 + 1 , − 5 + 1 2 ) = 1 2 | 6 + 2 5 + 4 5 = 4 2 | = 6 5 + 1 0 4 = 3 5 + 5 2

If the tangents drawn at the points O(0, 0) and <math> <mrow> <mi>P</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>+</mo> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> &nbsp;on the circle x2 + y2 – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

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

If the tangents drawn at the points O(0, 0) and $P (1 + \sqrt[]{5}, 2)$ on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

Option 1 -

$\frac{3 + \sqrt[]{5}}{2}$

Option 2 -

$\frac{4 + 2 \sqrt[]{5}}{2}$

Option 3 -

$\frac{5 + 3 \sqrt[]{5}}{2}$

Option 4 -

$\frac{7 + 3 \sqrt[]{5}}{2}$

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 1

Detailed Solution:

<div class="photo-widget-full"><div class="figure"><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1757994341phpxcJmyc.jpeg" alt="" width="202" height="146" loading="lazy"></picture></div></div> <math> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>−</mo> <mn>4</mn> <mi>y</mi> <mo>=</mo> <mn>0</mn> </mrow> </math> Centre (1, 2) r = <math> <mrow> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>  Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0⇒x + 2y = 0       ……… (i)Equation of PQ is  <math> <mrow> <mi>x</mi> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>+</mo> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <mn>2</mn> <mi>y</mi> <mo>−</mo> <mrow> <mo> (</mo> <mrow> <mo>|</mo> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mo>+</mo> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> <mo>−</mo> <mn>2</mn> <mrow> <mo> (</mo> <mrow> <mi>y</mi> <mo>+</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> </mrow> </math>  Solving (i) & (ii),   <math> <mrow> <mi>Q</mi> <mrow> <mo> (</mo> <mrow> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> <mo>, </mo> <mo>−</mo> <mfrac> <mrow> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>                <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mrow> <mo>|</mo> <mrow> <mfrac> <mrow> <mn>6</mn> <mo>+</mo> <mn>2</mn> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>4</mn> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>4</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>|</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> <mroot> <mrow> <mn>5</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>   </div></div>

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If the tangents drawn at the points O(0, 0) and $P (1 + \sqrt[]{5}, 2)$ on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

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If the tangents drawn at the points O(0, 0) and $P (1 + \sqrt[]{5}, 2)$ on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to