Question

The triangle formed by the normal to the curve $f\left(x\right)=x^2-ax+2a$ at the point $(2,4)$ and the coordinate axes lie in second quadrant. If its area is $2$ sq.units, then the sum of possible values of $a$ is

Answer 1

$y=x^2-ax+2a\Rightarrow \frac{dy}{dx}=2x-a$
Now point is $(2,4),$ so $\frac{dy}{dx}=4-a$

Equation of normal at $(2,4)$ is
$y-4=\frac{-1}{dy/dx}(x-2)\Rightarrow (a-4)(y-4)=(x-2)$
$\Rightarrow x-(a-4)y=18-4a$

$y-$intercept $:\frac{18-4a}{-(a-4)}=\frac{4a-18}{a-4}>0$
$\Rightarrow a\in (-\infty ,4)\cup (\frac{9}{2},\infty )\cdots \left(1\right)$

$x-$intercept $:18-4a<0$
$\Rightarrow a>\frac{9}{2}\cdots \left(2\right)$
$\left(1\right)\cap \left(2\right),$ we get
$a>\frac{9}{2}$

Now, $\frac{1}{2}\times \left|\frac{4a-18}{a-4}\right|\times |18-4a|=2$
$\Rightarrow (4a-18{)}^2=4(a-4)$
$\Rightarrow a=5(a\ne \frac{17}{4}\text{as}a>\frac{9}{2})$