Let one end of a focal chord of the parabola y^2=16x be (16,16).

Question

Let one end of a focal chord of the parabola ${y}^{2}=16x$ be $(16,16)$. If $P(\alpha ,\beta )$ divides this focal chord internally in the ratio $5:2$, then the minimum value of $\alpha +\beta $ is equal to:

Accepted Answer

Here $4a=16\Rightarrow a=4$. $(16,16)$ corresponds to $t_1=2$ since $(4t^2,8t)=(16,16)$. For a focal chord, $t_1t_2=-1\Rightarrow t_2=-\dfrac12$, giving the other end $B=(1,-4)$.

Dividing $A(16,16)$ and $B(1,-4)$ internally in ratio $5:2$ (from $A$ to $B$):

$$P=\left(\frac{5(1)+2(16)}{7},\ \frac{5(-4)+2(16)}{7}\right)=\left(\frac{37}{7},\frac{12}{7}\right),\qquad \alpha+\beta=7.$$

(Taking the chord in the other direction gives $\alpha+\beta=\dfrac{82}{7}+\dfrac{72}{7}=22$.) The minimum value is $7$.

Solutions · 1