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

NTA Abhyas 2020 · Mathematics · Ellipse

The length of the minor axis of the ellipse (5x - 10) ^ 2 + (5y +...

The length of the minor axis of the ellipse (5x−10){\left(5x - 10\right)}^{2} + {\left(5y + 15\right)}^{2} = \dfrac{{\left(3x - 4y + 7\right)}^{2}}{4} is

Options

  1. A103\dfrac{10}{3} units
  2. B103\dfrac{10}{\sqrt{3}} units
  3. C203\dfrac{20}{3} units
  4. D53\dfrac{5}{\sqrt{3}} units

Correct answer

D. 53\dfrac{5}{\sqrt{3}} units

Step-by-step solution

{\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {\left(\dfrac{1}{2}\dfrac{3x - 4y + 7}{5}\right)}^{2} is an ellipse, whose focus is (2,−3),\left(2, - 3\right), directix 3x−4y+7=03x - 4y + 7 = 0 and eccentricity is 12\dfrac{1}{2}
Length of ⊥⊥ from focus to directrix is 3×2−4×(−3)+75=5\dfrac{3 \times 2 - 4 \times \left(- 3\right) + 7}{5} = 5
ae−ae=5⇒2a−a2=5⇒a=103\dfrac{a}{e} - ae = 5 ⇒ 2a - \dfrac{a}{2} = 5 ⇒ a = \dfrac{10}{3}
So, length of major axis is 203\dfrac{20}{3}

and length of minor axis,~b=53b = \dfrac{5}{3}

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