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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If x^ 2 + y^ 2 = a^ 2 and x^ 2 16 + y^ 2 9 = 1 intersect at 4...

If x2+y2=a2x^{2} + y^{2} = a^{2} and x216+y29=1\dfrac{x^{2}}{16} + \dfrac{y^{2}}{9} = 1 intersect at 44 points P,Q,RP, Q, R and SS which form a square, then the area (in sq. units) of the square isIf x^ 2 + y^ 2 = a^ 2 and x^ 2 16 + y^ 2 9 = 1 intersect at 4 points P, Q, R and S which form a square, then the area

Options

  1. A14425\dfrac{144}{25}
  2. B254\dfrac{25}{4}
  3. C152\dfrac{15}{2}
  4. D57625\dfrac{576}{25}

Correct answer

D. 57625\dfrac{576}{25}

Step-by-step solution

If x^ 2 + y^ 2 = a^ 2 and x^ 2 16 + y^ 2 9 = 1 intersect at 4 points P, Q, R and S which form a square, then the area
Let,& P = \left( acos θ,\theta, asin \theta\right)<br />Then, \left( acos θ,\theta, asin \theta\right) lies on \dfrac{x^{2}}{16} + \dfrac{y^{2}}{9} =1
Also, \theta =45°45 ° because P,Q,R,SP, Q, R, S form a square
⇒a22×16+a22×9=1⇒ \dfrac{a^{2}}{2 \times 16} + \dfrac{a^{2}}{2 \times 9} =1
⇒a22(16+916×9)=1⇒a2=2×14425<br/>⇒ \dfrac{a^{2}}{2}\left(\dfrac{16 + 9}{16 \times 9}\right) =1 ⇒ a^{2} = \dfrac{2 \times 144}{25}<br />⇒ a = 1225\dfrac{12\sqrt{2}}{5}⇒ PQ = 245</p><p>Hence,theareaofthesquare=57625\dfrac{24}{5}</p><p>Hence, the area of the square=\dfrac{576}{25}</p></p>

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