Shortest Distance Between Lines and Ellipse Relation

 

❓ Question

Let the values of $𝑝$, for which the shortest distance between the lines

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

and

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

is

$$\frac{1}{\sqrt{6}}$$

be $𝑎,𝑏$ $(𝑎<𝑏)$.

Then the length of latus rectum of the ellipse

$$\frac{{𝑥}^2}{{𝑎}^2}+\frac{{𝑦}^2}{{𝑏}^2}=1$$

is equal to ________.

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🖼 Question Image

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✍️ Short Explanation

This problem combines:

👉 Shortest distance between skew lines
👉 Parameter-based quadratic equation
👉 Ellipse latus rectum formula.

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🔷 Step 1 — Write Lines in Vector Form

First line:

$$\frac{𝑥+1}{3}=\frac{𝑦}{4}=\frac{𝑧}{5}=𝑡$$

So:

$$\overrightarrow{𝑟}=(-1,0,0)+𝑡(3,4,5)$$

Direction vector:

$${\overrightarrow{𝑑}}_1=(3,4,5)$$

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Second line:

$$\overrightarrow{𝑟}=(𝑝,2,1)+𝜆(2,3,4)$$

Direction vector:

$${\overrightarrow{𝑑}}_2=(2,3,4)$$

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🔷 Step 2 — Formula for Shortest Distance

Shortest distance between skew lines:

$$𝐷=\frac{\mathrm{\mid }({\overrightarrow{𝑎}}_2-{\overrightarrow{𝑎}}_1)\cdot ({\overrightarrow{𝑑}}_1\times {\overrightarrow{𝑑}}_2)\mathrm{\mid }}{\mathrm{\mid }{\overrightarrow{𝑑}}_1\times {\overrightarrow{𝑑}}_2\mathrm{\mid }}$$

Given:

$$𝐷=\frac{1}{\sqrt{6}}$$

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🔷 Step 3 — Calculate Cross Product

$${\overrightarrow{𝑑}}_1\times {\overrightarrow{𝑑}}_2=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 3& 4& 5\\ 2& 3& 4\end{array}\mid$$
$$=(16-15)\widehat{𝑖}-(12-10)\widehat{𝑗}+(9-8)\widehat{𝑘}$$
$$=\widehat{𝑖}-2\widehat{𝑗}+\widehat{𝑘}$$

Magnitude:

$$\mathrm{\mid }{\overrightarrow{𝑑}}_1\times {\overrightarrow{𝑑}}_2\mathrm{\mid }=\sqrt{1+4+1}=\sqrt{6}$$

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🔷 Step 4 — Apply Distance Condition

Points on lines:

$${\overrightarrow{𝑎}}_1=(-1,0,0)$$
$${\overrightarrow{𝑎}}_2=(𝑝,2,1)$$

So:

$${\overrightarrow{𝑎}}_2-{\overrightarrow{𝑎}}_1=(𝑝+1,2,1)$$

Dot product:

$$(𝑝+1,2,1)\cdot (1,-2,1)$$
$$=𝑝+1-4+1$$
$$=𝑝-2$$

Thus:

$$\frac{\mathrm{\mid }𝑝-2\mathrm{\mid }}{\sqrt{6}}=\frac{1}{\sqrt{6}}$$
$$\mathrm{\mid }𝑝-2\mathrm{\mid }=1$$
$$𝑝=3,1$$

Hence:

$$𝑎=1,𝑏=3$$

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🔷 Step 5 — Ellipse Latus Rectum

Ellipse:

$$\frac{{𝑥}^2}{{𝑎}^2}+\frac{{𝑦}^2}{{𝑏}^2}=1$$

Since:

$$𝑏>𝑎$$

Major axis:

$$=3$$

Minor axis:

$$=1$$

Length of latus rectum:

$$\frac{2{𝑏}^2}{𝑎}$$
$$=\frac{2(1{)}^2}{3}$$
$$=\frac{2}{3}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Cross product calculation galat kar dena

❌ Modulus equation miss kar dena

❌ Ellipse mein major/minor axis ulta use kar lena

Remember:

$$\text{Larger denominator}\Rightarrow \text{major axis}$$

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✅ Final Answer

$$\boxed{{\displaystyle \frac{2}{3}}}$$

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📚 Related Topics

• Hyperbola Latus Rectum Right Angle Problem
• Derive Parabola Equation from Focus and Directrix
• Find Product of Focal Distances Hyperbola