The equation of the tangents to the ellipse `4x^(2)+3y^(2)=5`, which are parallel to the line y=3x+7 are

Options

A. `y=3x+-sqrt((155)/(3))`
B. `y=3x+-sqrt((155)/(12))`
C. `y=3x+-sqrt((95)/(12))`
D. none of these

Answer

B

Explanation

the equation of the given ellipse may be writen in the form
`(x^(2))/(5//4)+(y^(2))/(5//3)=1.` here `a^(2)=(5)/(4) ,b^(2)=(5)/(3)`
the slope of the given line is 3, therefore the slope of the parallel line is also 3.
the equations of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` are
`y=ms+- sqrt(a^(2)m^(2)+b^(2))`
in this case , we have
`m=3,a^(2)=(5)/(4),b^(2)=(5)/(3)`
So the required equations of the tangents are `y=3x+- sqrt((155)/(12)).`