Question

The tangent to the parabola $y=x^2+ax+1$ at the point of intersection of $y-$axis also touches the circle $x^2+y^2=r^2$ and no point of the parabola is below $x-$axis. Then

• A. the radius of circle when ${}^{\prime }{a}^{\prime }$ attains its maximum value is $\frac{1}{\sqrt{10}}$
• B. the slope of the tangent when radius of the circle is maximum is $0$
• C. area bounded by the tangent and the coordinate axes is minimum when $a=1$
• D. the minimum area bounded by the tangent and the coordinate axes is $\frac{1}{4}$ sq. unit

Answer 1

Answer: (B), (D)

the minimum area bounded by the tangent and the coordinate axes is $\frac{1}{4}$ sq. unit

$y=x^2+ax+1$
Since, no point of the parabola is below $x-$axis,
$\therefore D=a^2-4\le 0$
Therefore, maximum possible value of $a$ is $2.$
Equation of the parabola, when $a=2$, is
$y=x^2+2x+1$
It intersects $y-$axis at $(0,1).$
$\therefore$ Equation of the tangent at $(0,1)$ is
$y=2x+1$
Since, $y=2x+1$ touches the circle $x^2+y^2=r^2,$
$\therefore r=\frac{1}{\sqrt{5}}$

Equation of the tangent at $(0,1)$ to the parabola $y=x^2+ax+1$ is
$y-1=a(x-0)$
$\Rightarrow ax-y+1=0$
As it touches the circle,
$\therefore r=\frac{1}{\sqrt{a^2+1}}$
Radius is maximum when $a=0$
Therefore, equation of the tangent, when $a=0$ is $y=1$.
Hence, slope of the tangent is $0.$

Equation of tangent is $y=ax+1$.
Intercepts are $-\frac{1}{a}$ and $1.$
Therefore, area of the triangle bounded by tangent and the axes is
$A=\frac{1}{2}|-\frac{1}{a}\cdot 1|=\frac{1}{2|a|}$
Area is minimum, when $a=2$
And minimum area $=\frac{1}{4}$ sq. unit