Find the equation of tangents to the curve `4x^2-9y^2=1` which are parallel to `4y=5x+7.`

Explanation

Let m be the slope of the tangent of `4x^(2)-9y^(2)=1.` Then `m=("Slope of the line "4y=5x+7)=(5)/(4)`
We have `(x^(2))/(1//4)-(y^(2))/(1//9)=1`
where `a^(2)=(1)/(4),b^(2)=(1)/(9)`
The equations of the tangents are
`y=mx pm sqrt(a^(2)m^(2)-b^(2))`
`"or "y=(5)/(4)xpmsqrt((25)/(64)-(1)/(9))`
`"or "30x-24y pm sqrt(161)=0`
`"or "24y-30x=pmsqrt(161)`