Question

12)If p and p' are the length of the perpendiculars drawn from the points (+-root a2-b2,0) to the line x/a cosA + y/b sinA = 1, prove that: p x p' = b2.

Answer 1

Answer:

Kindly refer to below explanation.

Step-by-step explanation:

Given equation of line:

$\dfrac{x}{a}cos\theta+ \dfrac{y}{b}sin\theta =1\\OR\\\dfrac{x}{a}cos\theta+ \dfrac{y}{b}sin\theta-1=0$

Given points are:

$(\sqrt{a^2-b^2},0), (-\sqrt{a^2-b^2},0)$

Formula for distance between a point and a line is:

If the point is $(m, n)$ and equation of line is $Ax+By+C = 0$

Then, perpendicular distance between line and points is:

$Distance = \dfrac{|Am+Bn+C|}{\sqrt{A^2+B^2}}$

Here,

$A = \dfrac{x}{a}cos\theta\\B = \dfrac{y}{b}sin\theta\\C = -1$

For the first point:

$m = \sqrt{a^2-b^2}, n = 0$

By the above formula:

$p$ and $p'$ can be calculated as:

$p\times p' =\\\Rightarrow \dfrac{|\dfrac{cos\theta}{a}\times \sqrt{a^2-b^2}+ 0 -1|}{\sqrt{\dfrac{cos^2\theta}{a^2}+\dfrac{sin^2\theta}{b^2}}}\times \dfrac{|-\dfrac{cos\theta}{a}\times \sqrt{a^2-b^2}+ 0 -1|}{\sqrt{\dfrac{cos^2\theta}{a^2}+\dfrac{sin^2\theta}{b^2}}}\\\Rightarrow \dfrac{1-(\sqrt{a^2-b^2}\times \frac{cos\theta}{a})^2}{\dfrac{cos^2\theta}{a^2}+\dfrac{sin^2\theta}{b^2}}}$

Formula used:

$(x+y)(x-y) = x^2 - y^2$

$\Rightarrow \dfrac{\dfrac{a^2-a^2cos^2\theta+b^2cos^2\theta}{a^2}}{\dfrac{b^2cos^2\theta+a^2sin^2\theta}{a^2b^2}}\\\Rightarrow b^2(\dfrac{a^2sin^2\theta+b^2cos^2\theta}{a^2sin^2\theta+b^2cos^2\theta})\\\Rightarrow b^2$

(Using the identity:

$sin^2\theta +cos^2\theta = 1$)

(Hence provded)