Let the values of p, for which the shortest distance between the lines x+13=y4=z5 and r=(p…

Question

Let the values of $p$, for which the shortest distance between the lines $\dfrac{x+1}{3}=\dfrac{y}{4}=\dfrac{z}{5}$ and $\vec r=(p\hat i+2\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$ is $\dfrac{1}{\sqrt6}$, be $a,b$ ($a<b$). Then the length of the latus rectum of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is:

Accepted Answer

$\vec a-\vec b=(p+1,2,1)$ (line points $(-1,0,0)$ and $(p,2,1)$), directions $(3,4,5)$ and $(2,3,4)$:
$$(3,4,5)\times(2,3,4)=(1,-2,1).$$
$$d=\frac{|(p+1)-4+1|}{\sqrt6}=\frac{|p-2|}{\sqrt6}=\frac{1}{\sqrt6}\implies p=3\text{ or }1.$$
So $a=1,\ b=3$. Ellipse: $\dfrac{x^2}{1}+\dfrac{y^2}{9}=1$ (major axis along $y$).
$$\text{Latus rectum}=\frac{2a^2}{b}=\frac{2(1)}{3}=\frac23.$$

Solutions · 1