TS EAMCET · Maths · Ellipse
If \(x+\sqrt{3} y=3\) is the tangent to the ellipse \(2 x^2+3 y^2=k\) at a point \(P\) then the equation of the normal to this ellipse at \(\mathrm{P}\) is
- A \(5 x-2 \sqrt{3} y=1\)
- B \(x-\sqrt{3} y=2\)
- C \(x-\sqrt{3} y+1=0\)
- D \(3 x-\sqrt{3} y=1\)
Answer & Solution
Correct Answer
(D) \(3 x-\sqrt{3} y=1\)
Step-by-step Solution
Detailed explanation
\(x+\sqrt{3} y=3\)...(i) \(2 x^2+3 y^2=k\) Equation of tangent at \(\mathrm{P}\left(x_1, y_1\right)\) \(2 x x_1+3 y y_1=\mathrm{k}\)...(ii) If (i) and (ii) represents the same line…
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