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Mathematics Question on angle between two lines

The angle between the pair of lines (x2+y2)sin⁡2α=(x cos⁡θ−y sin⁡θ)2({{x}^{2}}+{{y}^{2}}){{\sin }^{2}}\alpha ={{(x\,\cos \theta -y\,\sin \theta )}^{2}} is :

A

θ\theta

B

2 θ2\,\theta

C

α\alpha

D

2 α2\,\alpha

Answer

2 α2\,\alpha

Explanation

Solution

Given pair of lines of (x2+y2)sin2α=(x cos⁡ θ−y sin⁡θ)2({{x}^{2}}+{{y}^{2}})si{{n}^{2}}\alpha ={{(x\,\cos \,\theta -y\,\sin \theta )}^{2}}
⇒\Rightarrow x2 sin⁡2α+y2 sin⁡2α=x2cos⁡2θ{{x}^{2}}\,{{\sin }^{2}}\alpha +{{y}^{2}}\,{{\sin }^{2}}\alpha ={{x}^{2}}{{\cos }^{2}}\theta
+y2 sin⁡2θ−2xy sin⁡θ cos⁡θ+{{y}^{2}}\,{{\sin }^{2}}\theta -2xy\,\sin \theta \,\cos \theta
⇒\Rightarrow x2(sin⁡2α−cos⁡2θ)+y2(sin⁡2α−sin⁡2θ){{x}^{2}}({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta )+{{y}^{2}}({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta )
+2xy  sin⁡θ cos⁡θ=0+2xy\,\,\sin \theta \,\cos \theta =0
⇒\Rightarrow x2(sin⁡2α−cos⁡2θ)+y2(sin⁡2α−sin⁡2θ){{x}^{2}}({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta )+{{y}^{2}}({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta )
+2(sin⁡θ cos⁡θ) xy=0+2(\sin \theta \,\cos \theta )\,xy=0
On comparing with ax2+by2+2hxy=0,a{{x}^{2}}+b{{y}^{2}}+2hxy=0,
We get, a=sin⁡2α−cos⁡2θ, b=sin⁡2α−sin⁡2θa={{\sin }^{2}}\alpha -{{\cos }^{2}}\theta ,\,b={{\sin }^{2}}\alpha -{{\sin }^{2}}\theta
and h=sin⁡ θ cos⁡θh=\sin \,\theta \,\cos \theta
Let θ\theta be the angle between the pair of lines. ∴\therefore
tan⁡θ=∣2 h2−aba+b∣\tan \theta =\left| \frac{2\,\sqrt{{{h}^{2}}-ab}}{a+b} \right|
=∣2sin⁡2 θ cos⁡2θ−(sin⁡2α−cos⁡2θ) ×(sin⁡2α−sin⁡2θ) sin⁡2α−cos⁡2θ+sin⁡2α−sin⁡2θ)∣=\left| \frac{2\sqrt{\begin{aligned} & {{\sin }^{2}}\,\theta \,{{\cos }^{2}}\theta -({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta ) \\\ & \times ({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta ) \\\ \end{aligned}}}{{{\sin }^{2}}\alpha -{{\cos }^{2}}\theta +{{\sin }^{2}}\alpha -{{\sin }^{2}}\theta )} \right|
=∣2sin⁡2θ cos⁡2θ−(sin⁡2α)2+sin⁡2α sin⁡2θ+sin⁡2α cos⁡2θ−sin⁡2θ cos⁡2θ −(−1−2sin⁡2α)∣=\left| \frac{2\sqrt{\begin{aligned} & {{\sin }^{2}}\theta \,{{\cos }^{2}}\theta -{{({{\sin }^{2}}\alpha )}^{2}}+{{\sin }^{2}}\alpha \\\ & {{\sin }^{2}}\theta +{{\sin }^{2}}\alpha \,{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \,{{\cos }^{2}}\theta \\\ \end{aligned}}}{-(-1-2{{\sin }^{2}}\alpha )} \right|
=∣2sin⁡2α(sin⁡2θ+cos⁡2θ)−(sin2α)2−cos⁡ 2  α∣=\left| \frac{2\sqrt{{{\sin }^{2}}\alpha ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )-{{(si{{n}^{2}}\alpha )}^{2}}}}{-\cos \,2\,\,\alpha } \right|
=∣2sin⁡2α(1−sin⁡2α)−cos⁡ 2α∣=\left| \frac{2\sqrt{{{\sin }^{2}}\alpha (1-{{\sin }^{2}}\alpha )}}{-\cos \,2\alpha } \right|
⇒\Rightarrow tan⁡θ=∣sin⁡ 2αcos⁡ 2α∣=tan⁡ 2α\tan \theta =\left| \frac{\sin \,2\alpha }{\cos \,2\alpha } \right|=\tan \,2\alpha
⇒\Rightarrow θ=2α\theta =2\alpha