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find the equation of the tangent and normal of the curve x^2-2y^2+4=0, perpendicular to 2x-y+7=0
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You, probably, know how to write an equation of the straight line perpendicular to 2x-y+7 = 0.
It is 2y+x+c = 0. It is the entire family of parallel lines depending on the coefficient "c".
Now, to get the tangent line to the given hyperbola, you need to find the coefficient "c" ("the line") from the system
= (1)
x + 2y + c = 0 (2)
under the condition that the system has one and only one solution.
In this way you will get the tangent line.
How to solve the system (1), (2)?
Very simple. Express x from (2) and substitute it into (1).
You will get a single quadratic equation for y.
The condition that it has one and only one real solution is vanishing its discriminant d = 0.
So, express that "d" via "c" and find "c" as a root of the equation d = 0.
For inspiration, see the lesson
Tangent lines and normal vectors to a hyperbola.
By the way, the curve x^2-2y^2+4=0 is a hyperbola.
Good luck!