JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the line \(y=m x\) and the ellipse \(2 x^{2}+y^{2}=1\) intersect at a ponit \(\mathrm{P}\) in the first quadrant. If the normal to this ellipse at \(P\) meets the co-ordinate axes at \(\left(-\frac{1}{3 \sqrt{2}}, 0\right)\) and \((0, \beta),\) then \(\beta\) is equal to
- A \(\frac{2}{\sqrt{3}}\)
- B \(\frac{2 \sqrt{2}}{3}\)
- C \(\frac{2 }{3}\)
- D \(\frac{\sqrt{2}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{2}}{3}\)
Step-by-step Solution
Detailed explanation
Any normal to the ellipse is \(\frac{\mathrm{x} \sec \theta}{\sqrt{2}}-\mathrm{y} \cos \mathrm{ec} \theta=-\frac{1}{2}\) \(\Rightarrow \frac{\mathrm{x}}{\left(\frac{-\cos \theta}{\sqrt{2}}\right)}+\frac{\mathrm{y}}{\left(\frac{\sin \theta}{2}\right)}=1\)…
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