∵ x 2 = 2y, p = 1 ∴ The distance from the focus to the alignment is equal to 1.
2 ; X 2-2Y 2 = 16, with 0: X 2-2Y 2 = 0 instead of 16.
Asymptote equation is x-√2y=0, x+√2y=0,
3. x 2/9-y 2/16 =1,the same as 2.
The asymptote equation of x 2/9-y 2/ 16 = 1 is 4x 3y = 0.
The asymptote equation of hyperbola y 2/25-x 2/9 =1is 5x 3y = 0.
4,
Substitute P(a.√(a-3)) into x 2+2y 2 = 9,
De: A 2+2 (A-3) = 9, that is, a 2+2A- 15 = 0.
Solution: a=-5 (truncation), a=3.
Then a=3