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
- A5 x-3 y+1=0
- Bx-3 y+ 5 =0
- C5 x-3 y+3=0
- 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