If p and q are the lengths of the perpendiculars from the origin on the lines, x cosec \alpha- y sec \alpha = k cot 2 \alpha and x sin \alpha + y cos \alpha = k sin2 \alpha respectively, then k2 is equal to :

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2021 · 31 Aug · Shift 1 · Q29 with Solution

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Correct answer: A

Use the distance of a point from a line

For a line $Ax+By+C=0,$ the perpendicular distance from the origin $(0,0)$ is $d=\frac{|C|}{\sqrt{A^2+B^2}}.$

First line and distance $p$

Given: $x\csc\alpha-y\sec\alpha=k\cot2\alpha$

Rewrite as: $x\csc\alpha-y\sec\alpha-k\cot2\alpha=0$

So, $A=\csc\alpha,\quad B=-\sec\alpha,\quad C=-k\cot2\alpha.$

Hence, $p=\frac{|k\cot2\alpha|}{\sqrt{\csc^2\alpha+\sec^2\alpha}}.$

Now,

=\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha\cos^2\alpha} =\frac{1}{\sin^2\alpha\cos^2\alpha}.$$ Therefore, $$\sqrt{\csc^2\alpha+\sec^2\alpha}=\frac{1}{\sin\alpha\cos\alpha}.$$ So, $$p=|k\cot2\alpha|\,\sin\alpha\cos\alpha.$$ Using $$\cot2\alpha=\frac{\cos2\alpha}{\sin2\alpha}, \qquad \sin2\alpha=2\sin\alpha\cos\alpha,$$ we get $$p=|k|\cdot \frac{|\cos2\alpha|}{|\sin2\alpha|}\cdot \sin\alpha\cos\alpha =|k|\cdot \frac{|\cos2\alpha|}{2}.$$ Thus, $$p^2=\frac{k^2\cos^22\alpha}{4}.$$ --- 3. **Second line and distance $q$** Given: $$x\sin\alpha+y\cos\alpha=k\sin2\alpha$$ Rewrite as: $$x\sin\alpha+y\cos\alpha-k\sin2\alpha=0$$ So, $$A=\sin\alpha,\quad B=\cos\alpha,\quad C=-k\sin2\alpha.$$ Hence, $$q=\frac{|k\sin2\alpha|}{\sqrt{\sin^2\alpha+\cos^2\alpha}}=|k\sin2\alpha|.$$ Therefore, $$q^2=k^2\sin^22\alpha.$$ --- 4. **Relate $k^2$ with $p^2$ and $q^2$** From above, $$4p^2=k^2\cos^22\alpha,$$ and $$q^2=k^2\sin^22\alpha.$$ Add them: $$4p^2+q^2=k^2(\cos^22\alpha+\sin^22\alpha)=k^2.$$ Hence, $$\boxed{k^2=4p^2+q^2}.$$ --- 5. **Check options** - A: $4p^2+q^2$ ✅ - B: $2p^2+q^2$ ❌ - C: $p^2+2q^2$ ❌ - D: $p^2+4q^2$ ❌ So the correct option is **A**.

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View written solutionFree

Correct answer: A

Use the distance of a point from a line

For a line $Ax+By+C=0,$ the perpendicular distance from the origin $(0,0)$ is $d=\frac{|C|}{\sqrt{A^2+B^2}}.$

First line and distance $p$

Given: $x\csc\alpha-y\sec\alpha=k\cot2\alpha$

Rewrite as: $x\csc\alpha-y\sec\alpha-k\cot2\alpha=0$

So, $A=\csc\alpha,\quad B=-\sec\alpha,\quad C=-k\cot2\alpha.$

Hence, $p=\frac{|k\cot2\alpha|}{\sqrt{\csc^2\alpha+\sec^2\alpha}}.$

Now,

=\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha\cos^2\alpha} =\frac{1}{\sin^2\alpha\cos^2\alpha}.$$ Therefore, $$\sqrt{\csc^2\alpha+\sec^2\alpha}=\frac{1}{\sin\alpha\cos\alpha}.$$ So, $$p=|k\cot2\alpha|\,\sin\alpha\cos\alpha.$$ Using $$\cot2\alpha=\frac{\cos2\alpha}{\sin2\alpha}, \qquad \sin2\alpha=2\sin\alpha\cos\alpha,$$ we get $$p=|k|\cdot \frac{|\cos2\alpha|}{|\sin2\alpha|}\cdot \sin\alpha\cos\alpha =|k|\cdot \frac{|\cos2\alpha|}{2}.$$ Thus, $$p^2=\frac{k^2\cos^22\alpha}{4}.$$ --- 3. **Second line and distance $q$** Given: $$x\sin\alpha+y\cos\alpha=k\sin2\alpha$$ Rewrite as: $$x\sin\alpha+y\cos\alpha-k\sin2\alpha=0$$ So, $$A=\sin\alpha,\quad B=\cos\alpha,\quad C=-k\sin2\alpha.$$ Hence, $$q=\frac{|k\sin2\alpha|}{\sqrt{\sin^2\alpha+\cos^2\alpha}}=|k\sin2\alpha|.$$ Therefore, $$q^2=k^2\sin^22\alpha.$$ --- 4. **Relate $k^2$ with $p^2$ and $q^2$** From above, $$4p^2=k^2\cos^22\alpha,$$ and $$q^2=k^2\sin^22\alpha.$$ Add them: $$4p^2+q^2=k^2(\cos^22\alpha+\sin^22\alpha)=k^2.$$ Hence, $$\boxed{k^2=4p^2+q^2}.$$ --- 5. **Check options** - A: $4p^2+q^2$ ✅ - B: $2p^2+q^2$ ❌ - C: $p^2+2q^2$ ❌ - D: $p^2+4q^2$ ❌ So the correct option is **A**.

Straight Lines and Pair of Straight Lines question

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