On the ellipse, $4x^2\, +\, 9y^2\, =\, 1$, the points at which the tangents are parallel to the line $8x = 9y$ are

We have ${ 4x }^{ 2 }+9{ y }^{ 2 }=1$ ...(1)

Differentiating w.r.t x, we get

$\displaystyle 8x+18y\frac { dy }{ dx } =0\Rightarrow \frac { dy }{ dx } =-\frac { 4x }{ 9y }$

The tangent at point $\left( x,y \right)$ will be parallel to the line $8x=9y$ if

$\displaystyle \frac { -4x }{ 9y } =\frac { 8 }{ 9 } \Rightarrow x=-2y$

Substituting this in (1), we get

$\displaystyle 4{ \left( -2y \right) }^{ 2 }+{ 9y }^{ 2 }=1\Rightarrow { 25y }^{ 2 }=1\Rightarrow y=\pm \frac { 1 }{ 5 }$

Thus the point where the tangent are parallel to $8x=9y$ are

$\displaystyle \left( \frac { -2 }{ 5 } ,\frac { 1 }{ 5 } \right)$ and $\displaystyle \left( \frac { 2 }{ 5 } ,\frac { -1 }{ 5 } \right)$