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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
A point on the ellipse, \(4x^2 + 9y^2 = 36\), where the normal is parallel to the line, \(4x -2y-5 = 0\) , is
- A \(\left( {\frac{9}{5},\frac{8}{5}} \right)\)
- B \(\left( {\frac{8}{5},-\frac{9}{5}} \right)\)
- C \(\left( {-\frac{9}{5},\frac{8}{5}} \right)\)
- D \(\left( {\frac{8}{5},\frac{9}{5}} \right)\)
Answer & Solution
Correct Answer
(C) \(\left( {-\frac{9}{5},\frac{8}{5}} \right)\)
Step-by-step Solution
Detailed explanation
Given ellipse is \(4{x^2} + 9{y^2} = 36\) \( \Rightarrow \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1\) Normal at the point is parallel to the line \(4x + 2y - 5 = 0\) Slope of normal \(=2\) slope of tangent \( = \frac{{ - 1}}{2}\) Point of contact to ellipse…
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