Let one end of a focal chord of the parabola | vedclass

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(B) The equation of the parabola is $y^{2}=16x$,which is of the form $y^{2}=4ax$ with $a=4$.Let the coordinates of point $A$ be $(at_{1}^{2}, 2at_{1})

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The equation of the parabola is $y^{2}=16x$,which is of the form $y^{2}=4ax$ with $a=4$.Let the coordinates of point $A$ be $(at_{1}^{2}, 2at_{1}) = (16, 16)$.Thus,$2(4)t_{1} = 16 \Rightarrow t_{1}=2$.Since $AB$ is a focal chord,the product of the parameters of its endpoints is $t_{1}t_{2}=-1$.Therefore,$2t_{2}=-1 \Rightarrow t_{2}=-\frac{1}{2}$.The coordinates of point $B$ are $(at_{2}^{2}, 2at_{2}) = (4(-\frac{1}{2})^{2}, 2(4)(-\frac{1}{2})) = (1, -4)$.Point $P(\alpha, \beta)$ divides the chord $AB$ internally in the ratio $5:2$.Case $1$: $P$ divides $AB$ in ratio $5:2$ (from $A$ to $B$):$\alpha = \frac{5(1) + 2(16)}{5+2} = \frac{37}{7}$,$\beta = \frac{5(-4) + 2(16)}{5+2} = \frac{12}{7}$.$\alpha+\beta = \frac{37+12}{7} = \frac{49}{7} = 7$.Case $2$: $P$ divides $BA$ in ratio $5:2$ (from $B$ to $A$):$\alpha = \frac{5(16) + 2(1)}{5+2} = \frac{82}{7}$,$\beta = \frac{5(16) + 2(-4)}{5+2} = \frac{72}{7}$.$\alpha+\beta = \frac{82+72}{7} = \frac{154}{7} = 22$.The minimum value of $\alpha+\beta$ is $7$.

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:

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