AP EAMCET · Maths · Ellipse
Let the eccentricity of the ellipse \(2 x^2+a y^2-8 x-2 a y\) \(+(8-a)=0\) be \(\frac{1}{\sqrt{3}}\). If the major axis of this ellipse is parallel to Y-axis, then the equation of the tangent to this ellipse with slope 1 is
- A \(x-y-1 \pm \sqrt{5}=0\)
- B \(x-y-3 \pm \sqrt{5}=0\)
- C \(x-y-3 \pm \sqrt{\frac{10}{3}}=0\)
- D \(x-y-1 \pm \sqrt{\frac{10}{}}=0\)
Answer & Solution
Correct Answer
(D) \(x-y-1 \pm \sqrt{\frac{10}{}}=0\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { } 2 x^2+a y^2-8 x-2 a y+8-a=0.....(i) \\ & \Rightarrow\left(2 x^2-8 x\right)+\left(a y^2-2 a y\right)=a-8 \\ & \Rightarrow \frac{(x-2)^2}{a}+\frac{(y-1)^2}{2}=1 \end{aligned} \] We have \(b^2=a^2\left(i-e^2\right)\) and given that major axis is…
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