Let the values of p, for which the shortest distance between the lines and is , be a, b, (a < b). Then the length of the latus rectum of the ellipse is : [2025]

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Solution

Q.
Let the values of p, for which the shortest distance between the lines $\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$ and $\vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ is $\frac{1}{\sqrt{6}}$ , be a, b, (a < b). Then the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is : [2025]

1 9

2 $\frac{2}{3}$

3 18

4 $\frac{3}{2}$

Ans.
(2)

We have lines $\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$

$\vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$

where, $\vec{a} = - \hat{i} + 0 \hat{j} + 0 \hat{k}$

$\vec{b} = p \hat{i} + 2 \hat{j} + \hat{k} then \vec{a} - \vec{b} = (- 1 - p) \hat{i} - 2 \hat{j} - \hat{k}$

$\vec{p} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} and \vec{q} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$

Then, $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix} |$

$\Rightarrow \vec{p} \times \vec{q} = \hat{i} (16 - 15) - \hat{j} (12 - 10) + \hat{k} (9 - 8)$

$= \hat{i} - 2 \hat{j} + \hat{k}$

Now, shortest distance $=$ $| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$

$\Rightarrow \frac{1}{\sqrt{6}} = \frac{| (- p — 1) + 4 - 1 |}{\sqrt{1 + 4 + 1}} \Rightarrow$ $| - p + 2 |$ $= 1$

$\Rightarrow$ p = 3 and 1, then a = 1 and b = 3 ( $∵$ a < b)

So, length of latus rectum of ellipse $\frac{x^{2}}{1^{2}} + \frac{y^{2}}{3^{2}} = 1$ is $\frac{2 a^{2}}{b} = \frac{2 \times 1}{3} = \frac{2}{3}$ .

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