Question:

If the largest to the parabola y²=x at a point (α, β), (β>0) is also a tangent to the ellipse x²+2y²=1, then α to:

√2+1

√2-1;

2√2-1;

2√2+1

T:y(β)=1/2(x+β²/2)2yβ=x+β²/2y=(1/2β)x+β/2m=1/2β,C=β/2β²/4=±√1/4β²+1/2β²/4=1/4β²+1/2→β²/4=1+2β²/4β²→β⁴-2β²-1=0→(β²-1)²=2→β²=√2+1