The point in the first quadrant of the ellipse x2/25+y2/144=1 at which tangents makes equal angles with the axes is
a) (2,144/3)
b)25/13,144/13
c)-25/13,144/13
d)-25/13,-144/13
Answers
Given : ellipse x²/25 + y²/144 = 1
To Find : the point in the first quadrant at which the tangent makes equal angles with the axes
a) ( 2,144/3)
b) (25/13,144/13)
c) (-25/13,144/13)
d) (-25/13,-144/13)
Solution:
x²/25 + y²/144 = 1
=> 2x/25 + (2y/144) dy/dx = 0
=> x/25 + (y/144) dy/dx = 0
=> (y/144) dy/dx = -x/25
=> dy/dx = - 144x/25y
tangent makes equal angles with the axes
and Quadrant is 1st
hence slope = - 1
=> - 1 = - 144x/25y
=> 25y = 144x
=> y = 144x/25
x²/25 + y²/144 = 1
=> x²/25 + (144x/25)²/144 = 1
=> x²/25 + 144x²/25² = 1
=> 25x² + 144x² = 25²
=> 169x² = 25²
=> x² = 25²/169
=> x² = 25²/13²
=> x = 25/13 as 1st Quadrant
y = 144x/25 = 144/13
(25/13,144/13)
option B is correct
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