The foci of a hyperbola coincide with the foci of the ellipse `(x^(2))/(25)+(y^(2))/(9)=1`. If the eccentricity of the hyperbola is `2`, then the equation of the tangent of this hyperbola passing through the point `(4,6)` is

Options

A. `2x-y-2=0`
B. `3x-2y=0`
C. `2x-3y+10=0`
D. `x-2y+8=0`

Explanation

For the ellipse `(x^(2))/(25)+(y^(2))/(9)=1`, we have `a=5`, `b=3`.
`:.e=sqrt(1-(9)/(25))=(4)/(5)`
So, the coordinates of the foci are `(+-4,0)`. These are also foci of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` whose eccentricity is `2`.
`:.ae=4impliesa=2`
Now, `b^(2)=a^(2)(e^(2)-1)impliesb^(2)=4(4-1)=12`
So, the equation of the hyperbola is `(x^(2))/(4)-(y^(2))/(12)=1`.
Clearly , point `(4,6)` lies on it. The equation of tangent to this hyperbola at `(4,6)` is
`(4x)/(4)-(6y)/(12)=1` or, `2x-y=2`