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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2021 · 31 August · Morning Shift · Mathematics · Ellipse

The line 12xcosθ + 5ysinθ = 60 is tangent to which of the following...

The line 12xcosθ+5ysinθ=6012xcos\theta + 5ysin\theta = 60 is tangent to which of the following curves ?

Options

  1. Ax2+y2=30{x}^{2} + {y}^{2} = 30
  2. B144x2+25y2=3600144{x}^{2} + 25{y}^{2} = 3600
  3. Cx2+y2=169{x}^{2} + {y}^{2} = 169
  4. D25x2+12y2=360025{x}^{2} + 12{y}^{2} = 3600

Correct answer

B. 144x2+25y2=3600144{x}^{2} + 25{y}^{2} = 3600

Step-by-step solution

Given that 12xcosθ+5ysinθ=6012xcos\theta + 5ysin\theta = 60

Divide the above equation with 6060 on both sides

xcosθ5+ysinθ12=1…(i)\dfrac{xcos\theta}{5} + \dfrac{ysin\theta}{12} = 1 \ldots \left(i\right)

Let ellipse x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1

⇒⇒ Equation of tangent at point (acosθ,bsinθ)( acos\theta, bsin\theta )
xacosθa2+ybsinθb2=1\dfrac{xacos\theta}{{a}^{2}} + \dfrac{ybsin\theta}{{b}^{2}} = 1

xcosθa+ysinθb=1…\dfrac{xcos\theta}{a} + \dfrac{ysin\theta}{b} = 1 \ldots\left(ii\right)</p><p>Usingequation(i)</p><p>Using equation \left(i\right) and \left(ii\right)⇒a=5⇒ a = 5
b=12b = 12
Equation of ellipse is x225+y2144=1\dfrac{{x}^{2}}{25} + \dfrac{{y}^{2}}{144} = 1

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