Question

Let $y=f\left(x\right)$ be a parabola of the form $y=x^2+ax+1$ such that no point of the parabola is below $x-$axis. If its tangent at the point of intersection with $y-$axis also touches the circle $x^2+y^2=r^2$, then minimum area bounded by the tangent and the coordinate axes(in sq.units) is

Answer 1

Equation of parabola is $y=x^2+ax+1$
It intersects $y-$axis at $(0,1).$
Equation of the tangent at $(0,1)$ to the parabola $y=x^2+ax+1$ is $\frac{y+1}{2}=\frac{a}{2}(x+0)+1$
i.e. $ax-y+1=0$
So, intercepts are $\frac{-1}{a}$ and $1$
$\therefore$ Area of the triangle bounded by tangent and the axes $=\frac{1}{2}|-\frac{1}{a}\cdot 1|=\frac{1}{2|a|}$
It is minimum when $a$ will be maximum.
Since no point of the parabola is below $x-$axis.
$\therefore a^2-4\le 4$
$\therefore$ maximum value of $a$ is $2.$
$\therefore$minimum area $=\frac{1}{4}$ sq.units