Let the function f(x)=2x^2-_e x, x0, be decreasing in (0,a) and increasing in (a,4).

Question

Let the function $f(x)=2x^2-\log_e x$, $x>0$, be decreasing in $(0,a)$ and increasing in $(a,4)$. A tangent to the parabola $y^2=4ax$ at a point $P$ on it passes through the point $(8a,8a-1)$ but does not pass through the point $\left(-\dfrac1a,0\right)$. If the equation of the normal at $P$ is $\dfrac x\alpha+\dfrac y\beta=1$, then $\alpha+\beta$ is equal to \_\_\_\_\_\_\_\_.

Answer: 45

Accepted Answer

$f'(x)=4x-\frac1x=0\Rightarrow a=\frac12$. Parabola: $y^2=2x$.

Tangent at $(x_1,y_1)$: $yy_1=x+x_1$, with $y_1^2=2x_1$. Through $(4,3)$: $3y_1=4+\frac{y_1^2}2\Rightarrow y_1^2-6y_1+8=0\Rightarrow y_1=2\text{ or }4$.

- $y_1=2,x_1=2$: tangent $2y=x+2$, which passes through $(-2,0)$ --- excluded.
- $y_1=4,x_1=8$: tangent $4y=x+8$, at $x=-2$ gives $y=\frac32\ne0$ --- valid.

Normal at $(8,4)$ (slope $-y_1=-4$): $y-4=-4(x-8)\Rightarrow4x+y=36\Rightarrow\dfrac x9+\dfrac y{36}=1$.

$$\alpha+\beta=9+36=45.$$

Solutions · 1