Test Index

NCERT Class XI Mathematics - Straight Lines - Solutions

© examsnet.com
Question : 76 of 74
Marks: +1, -0
Prove that the product of the lengths of the perpendiculars drawn from the points (a2−b2,0)(\sqrt{a^2-b^2},0) and (−a2−b2,0)(-\sqrt{a^2-b^2},0) to the line xa\frac{x}{a} cos θ + yb\frac{y}{b} sin θ = 1 is b2b^2
Solution:  
The given equation of line is
xa\frac{x}{a} cos θ + yb\frac{y}{b} sin θ = 1 ... (i)
Now distance of (i) from the point (a2−b2,0)(\sqrt{a^2-b^2},0)
= ∣a2−b2acos⁡θ−1∣(cos⁡θa)2+(sin⁡θb)2\frac{\left|\frac{\sqrt{a^2-b^2}}{a}\cos\theta-1\right|}{\sqrt{\left(\frac{\cos\theta}{a}\right)^2+\left(\frac{\sin\theta}{b}\right)^2}}
And distance of (i) from the point (−a2−b2,0)(-\sqrt{a^2-b^2},0)
= ∣−a2−b2acos⁡θ−1∣(cos⁡θa)2+(sin⁡θb)2\frac{\left|-\frac{\sqrt{a^2-b^2}}{a}\cos\theta-1\right|}{\sqrt{\left(\frac{\cos\theta}{a}\right)^2+\left(\frac{\sin\theta}{b}\right)^2}}
Now, product of lengths of these two perpendiculars
=
∣(−a2−b2acos⁡θ−1)(−a2−b2acos⁡θ−1)∣cos⁡2θa2+sin⁡2θb2\frac{\left|\left(-\frac{\sqrt{a^2-b^2}}{a}\cos\theta-1\right)\left(-\frac{\sqrt{a^2-b^2}}{a}\cos\theta-1\right)\right|}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}
= ∣a2−b2a2cos⁡2θ−1∣b2cos⁡2θ+a2cos⁡2θa2b2\frac{\left|\frac{a^2-b^2}{a^2}\cos^2\theta-1\right|}{\frac{b^2\cos^2\theta+a^2\cos^2\theta}{a^2b^2}} = ∣(a2cos⁡2θ−b2cos⁡2θ)−a2∣⋅a2b2(b2cos⁡2θ+a2sin⁡2θ)a2\frac{\left|(a^2\cos^2\theta-b^2\cos^2\theta)-a^2\right|\cdot a^2b^2}{(b^2\cos^2\theta+a^2\sin^2\theta)a^2}
= ∣a2(cos⁡2θ−1)−b2cos⁡2θ∣⋅b2a2sin⁡2θ+b2cos⁡2θ\frac{\left|a^2(\cos^2\theta-1)-b^2\cos^2\theta\right|\cdot b^2}{a^2\sin^2\theta+b^2\cos^2\theta} = ∣−a2sin⁡2θ−b2cos⁡2θ∣⋅b2a2sin⁡2θ+b2cos⁡2θ\frac{\left|-a^2\sin^2\theta-b^2\cos^2\theta\right|\cdot b^2}{a^2\sin^2\theta+b^2\cos^2\theta}
= ∣−a2sin⁡2θ+b2cos⁡2θa2sin⁡2θ+b2cos⁡2θ∣b2\left|-\frac{a^2\sin^2\theta+b^2\cos^2\theta}{a^2\sin^2\theta+b^2\cos^2\theta}\right|b^2 = ∣−1∣b2\left|-1\right|b^2 = b2b^2
Hence proved.
© examsnet.com
Go to Question:
Prev Question
Next Question