Question:
Show that the product of the lengths of the perpendiculars drawn from the points $(\pm c,0)$ to the straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
My attempt:
Let, the length of the perpendicular on drawn upon $bx\cos\theta+ay\sin\theta-ab=0$ from $(c,0)$ is $d_1$ and the length of that drawn upon the same from $(-c,0)$ is $d_2$. Now,
$$d_1=\frac{|bc\cos\theta+0-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1=\frac{|bc\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
Again,
$$d_2=\frac{|-bc\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
Now,
$$d_1d_2=\frac{|-(bc\cos\theta-ab)(bc\cos\theta+ab)|}{{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2(1-\cos^2\theta)}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2-a^2\cos^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{a^2-(a^2-b^2)\cos^2\theta}$$
$$\implies d_1d_2=|\frac{b^2(a^2-c^2\cos^2\theta)}{a^2-(a^2-b^2)\cos^2\theta}|$$
[$a^2-(a^2-b^2)\cos^2\theta$ is positive, so bringing it inside the modulus sign doesn't make a difference]
$$\implies d_1d_2=|\frac{b^2(a^2-(a^2-b^2)\cos^2\theta)}{(a^2-(a^2-b^2)\cos^2\theta)}|$$
[Given, $a^2=b^2+c^2\implies c^2=a^2-b^2$]
$$d_1d_2=|b^2|$$(shown?)
The question asked me to show that $d_1d_2=b^2$, but I showed them that $d_1d_2=|b^2|$. So, is there something wrong with my process, or is there something wrong with the question?
Best Answer
Comment
Btw, the product of the length of the perpendiculars drawn from the focal points of an ellipse $(\pm c,0)$ to the tangent straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
Can be also derived from Newton's ellipse canonical form.