If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos heta - y\sin heta = k\cos 2 heta$ and $x\sec heta + y\, ext{cosec}\, heta = k$, respectively, prove that $p^2 + 4q^2 = k^2$.

Question

If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos	heta - y\sin	heta = k\cos 2	heta$ and $x\sec	heta + y\,	ext{cosec}\,	heta = k$, respectively, prove that $p^2 + 4q^2 = k^2$.

Accepted Answer

> [!TLDR]
> The problem reduces to computing perpendicular distances from the origin to two given lines, then simplifying $p^2 + 4q^2$ using trigonometric identities to obtain $k^2$.

## Concept and Intuition

When a problem asks for the perpendicular distance from the origin to a line, the standard formula is your friend: for a line $ax + by + c = 0$, the distance from $(0,0)$ is $\frac{|c|}{\sqrt{a^2 + b^2}}$. Here, both lines are given in forms that look different — one has $\cos	heta$ and $\sin	heta$, the other has $\sec	heta$ and $\csc	heta$. The trick is to rewrite each line in the standard form, compute $p$ and $q$, then combine them.

The result $p^2 + 4q^2 = k^2$ is neat because it's independent of $	heta$ — the trigonometric terms cancel out completely. That's the sign of a well-constructed identity.

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## Step-by-Step Solution

**1. First line: $x\cos	heta - y\sin	heta = k\cos 2	heta$**

Rewrite in standard form $ax + by + c = 0$:
$$x\cos	heta - y\sin	heta - k\cos 2	heta = 0$$

Here $a = \cos	heta$, $b = -\sin	heta$, $c = -k\cos 2	heta$.

The perpendicular distance $p$ from the origin is:
$$p = \frac{|c|}{\sqrt{a^2 + b^2}} = \frac{|-k\cos 2	heta|}{\sqrt{\cos^2	heta + \sin^2	heta}}$$

Since $\cos^2	heta + \sin^2	heta = 1$, the denominator is $1$. Also, $k$ is presumably positive (length), so:
$$p = |k\cos 2	heta|$$

> [!NOTE]
> The absolute value matters for distance, but since we'll square $p$ later, we can drop the absolute sign: $p^2 = k^2\cos^2 2	heta$.

**2. Second line: $x\sec	heta + y\,	ext{cosec}\,	heta = k$**

Rewrite in standard form:
$$x\sec	heta + y\,	ext{cosec}\,	heta - k = 0$$

Here $a = \sec	heta$, $b = 	ext{cosec}\,	heta$, $c = -k$.

The perpendicular distance $q$ from the origin is:
$$q = \frac{|c|}{\sqrt{a^2 + b^2}} = \frac{|-k|}{\sqrt{\sec^2	heta + 	ext{cosec}^2	heta}}$$

So $q = \frac{k}{\sqrt{\sec^2	heta + 	ext{cosec}^2	heta}}$.

**3. Simplify the denominator for $q$**

Recall $\sec^2	heta = \fra

Q.If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos\theta - y\sin\theta = k\cos 2\theta$ and $x\sec\theta + y\,\text{cosec}\,\theta = k$ , respectively, prove that $p^2 + 4q^2 = k^2$ .

Concept understanding — Distance From Point To Line

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