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

Options

A. `24y-30x=17`
B. `30y-24x=+-sqrt(161)`
C. `24y-30x=+-sqrt(161)`
D. none of these

Explanation

Let `m` be the slope of the tangent to `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` or, `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, where `a^(2)=(1)/(4)`, `b^(2)=(1)/(9)`
The equations of the tangents are
`y=mx+-sqrt(a(2)m^(2)-b^(2))`
`impliesy=(5)/(4)x+-sqrt((25)/(64)-(1)/(9))`
`implies30x-24y+-sqrt(161)=0implies24y-30x=+-sqrt(161)`