Question

( 4 x^{2}+9 y^{2}=1 )
Differentiating both sides.
[
begin{array}{l}
8 x+18 y frac{d y}{d x}=0
frac{d y}{d x}=frac{-8 x}{18 y}=frac{-4 x}{9 y}
end{array}
]
( therefore ) slope of tangent at point ( (x, y), m_{T}=left[frac{d y}{d x}right] )
[
=-frac{4 x}{9 y}
]
given line ( : 8 x=9 y )
[
begin{aligned}
therefore quad 8 &=9 frac{d y}{d x}
frac{d y}{d x} &=frac{8}{9}
end{aligned}
]
( Rightarrow quad m_{T}=frac{8}{9} )
( r: ) line & tangew
[
begin{array}{l}
frac{8}{80,}=-frac{4 x}{9 y} Rightarrow 4 x=-8 y
x=-2 y
end{array}
]
since this point lies on ginen curare ( 4(-2 y)^{2}+9 y^{2}=1 Rightarrow y=pm frac{1}{5} ln x=pm frac{2}{5} )

# The points on the curve 4x2 +9y2 = 1, at which the tangents are parallel to the line 8x = 9y are

Solution