AP EAMCET · Maths · Ellipse
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
- A \(\sqrt{5} x-3 y+1=0\)
- B \(x-3 y+\sqrt{5}=0\)
- C \(\sqrt{5} x-3 y+3=0\)
- D \(\sqrt{5} x-3 y+9=0\)
Answer & Solution
Correct Answer
(D) \(\sqrt{5} x-3 y+9=0\)
Step-by-step Solution
Detailed explanation
Equation of given ellipse is \[ 4 x^2+9 y^2=36 \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1 \] Now, coordinate of end of the latus rectum lying in the second quadrant is \(P\left(-\sqrt{5}, \frac{4}{3}\right)\). So, the equation of tangent at point \(P\) is…
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