Analytic Geometry – Problem 11: Find

For ellipse $\dfrac{x^2}{2}+y^2=1$, tangents from $P(x_0,y_0)$ touch at $A,B$. Slopes are $k_1,k_2,k_3,k_4$ for $OA,OB,PA,PB$, and $k_1k_2=\dfrac14$. Find $\min(5x_0-3y_0+k_3k_4)$.

For ellipse $\dfrac{x^2}{2}+y^2=1$, the tangent at $(x_t,y_t)$ is $$\frac{x_tx}{2}+y_ty=1,$$ so its slope is proportional to $1/(2\,y_t/x_t)$. Therefore (under the slope notation in the question), $$k_3=\frac{1}{2k_1},\qquad k_4=\frac{1}{2k_2}.$$ Hence $$k_3k_4=\frac{1}{4k_1k_2}=\frac{1}{4\cdot\frac14}=1.$$

So we only need to minimize $$5x_0-3y_0+k_3k_4=(5x_0-3y_0)+1.$$ For this ellipse, the chord of contact from $P(x_0,y_0)$ is $$\frac{x_0}{2}x+y_0y=1.$$ Combining with $k_1k_2=\dfrac14$, the corresponding point $P$ satisfies $$x_0^2-y_0^2=1,\qquad x_0>0,\ y_0>0.$$ Set $x_0=\cosh t,\ y_0=\sinh t$ $(t\ge0)$. Then $$5x_0-3y_0=5\cosh t-3\sinh t=4\cosh\!\left(t-\operatorname{artanh}\frac35\right)\ge4.$$ Equality holds when $\tanh t=\dfrac35$, i.e. $(x_0,y_0)=\left(\dfrac54,\dfrac34\right)$. Therefore $$\min(5x_0-3y_0+k_3k_4)=4+1=5.$$

$$\min(5x_0-3y_0+k_3k_4)=5.$$