Answer:
Kindly refer to below explanation.
Step-by-step explanation:
Given equation of line:
[tex]\dfrac{x}{a}cos\theta+ \dfrac{y}{b}sin\theta =1\\OR\\\dfrac{x}{a}cos\theta+ \dfrac{y}{b}sin\theta-1=0[/tex]
Given points are:
[tex](\sqrt{a^2-b^2},0), (-\sqrt{a^2-b^2},0)[/tex]
Formula for distance between a point and a line is:
If the point is [tex](m, n)[/tex] and equation of line is [tex]Ax+By+C = 0[/tex]
Then, perpendicular distance between line and points is:
[tex]Distance = \dfrac{|Am+Bn+C|}{\sqrt{A^2+B^2}}[/tex]
Here,
[tex]A = \dfrac{x}{a}cos\theta\\B = \dfrac{y}{b}sin\theta\\C = -1[/tex]
For the first point:
[tex]m = \sqrt{a^2-b^2}, n = 0[/tex]
By the above formula:
[tex]p[/tex] and [tex]p'[/tex] can be calculated as:
[tex]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}}}[/tex]
Formula used:
[tex](x+y)(x-y) = x^2 - y^2[/tex]
[tex]\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[/tex]
(Using the identity:
[tex]sin^2\theta +cos^2\theta = 1[/tex])
(Hence provded)
