Question

fnd the normal line of to the hyperbola 3x^2-4y^2=8 parallel to the line 2x-3y=7

Answer 1

rewrite equation of hyperbola as \[3x^2/8 -4y^2/8=1\] (divide by 8 on both sides)
\[a^2=8/3 \] \[b^2= 2\]

Answer 2

Now slope of normal is equal to slope of given line 2x-3y=7----> m=2/3

Answer 3

apply tangency condition,
\[a^2m^2-b^2=c^2\] find c. substitute in line equation to get points of tangency \[(-a^2m/c, -b^2/c)\]
substitute those points in \[y-y1=m(x-x1)\]