TS EAMCET · Maths · Ellipse
If \(x+2 y+k=0, k>0\) is a tangent to the ellipse \(2 x^2+y^2=2\), then the equation of the normal to the given ellipse at \(\left(\frac{1}{\sqrt{2}}, \frac{k}{3}\right)\), is
- A \(\sqrt{2} x-2 y+1=0\)
- B \(3 \sqrt{2} x-y-2=0\)
- C \(2 \sqrt{2} x-5 y+3=0\)
- D \(\sqrt{2} x+3 y-4=0\)
Answer & Solution
Correct Answer
(A) \(\sqrt{2} x-2 y+1=0\)
Step-by-step Solution
Detailed explanation
Given ellipse \(2 x^2+y^2=2 \text { or } x^2+\frac{y^2}{2}=1\) Here, \(\quad a^2=1, b^2=2\) Equation of tangent \(x+2 y+k=0 \text { or } 2 y=-x-k\) or \(\quad y=-\frac{x}{2}-\frac{k}{2}\) Here, \(c=\frac{-k}{2}, m=-\frac{1}{2}\) From the condition of tangent…
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