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AP EAMCET201822 Apr 2018Evening ShiftMathematicsEllipseActual

The equation of the tangent of the ellipse 4 x^2+9 y^2=36 at the end of the latusrectum lying in the second quadrant, is

Options

  1. A5 x-3 y+1=0
  2. Bx-3 y+ 5 =0
  3. C5 x-3 y+3=0
  4. D5 x-3 y+9=0

Correct answer

D. 5 x-3 y+9=0

Step-by-step solution

Equation of given ellipse is 4 x^2+9 y^2=36 x^2 9 + y^2 4 =1 Now, coordinate of end of the latus rectum lying in the second quadrant is P (- 5 , 4 3 ) . So, the equation of tangent at point P is array rlrl -4 5 x+12 y & =36 & 5 x-3 y+9 & =0 array

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