Tangents are drawn to the hyperbola `3x^2-2y^2=25` from the point `(0,5/2)dot` Find their equations.

Correct Answer - `y=(5)/(2)pm(3)/(2)x`
The hyperbola is
`3x^(2)-2y^(2)=25`
`"or "(x^(2))/(25//3)-(y^(2))/(25//2)=1" (1)"`
Tangent to the hyperbola having slope m is
`y=mx+sqrt(a^(2)m^(2)-b^(2))" (2)"`
For the given hyperbola =, the tangent having slope m is
`y=mx+sqrt((25)/(3)m^(2)-(25)/(2))`
It passes through (0, 5/2). Therefore,
`(25)/(4)=(25)/(3)m^(2)-(25)/(2)`
`"or "(1)/(4)=(2m^(2)-3)/(6)`
`"or "m^(2)=(9)/(4)`
`"or "m= pm (3)/(2)`
Therefore, the equation of tangents are
`y-(5)/(2)=pm(3)/(2)(x-0)`
`"or "y=(5)/(2)pm(3)/(2)x`