Question

The line perpendicular to $x-$axis and passing through $A(a\cos \theta ,0)$ meets the striaght line $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ and $x\cos \theta +y\sin \theta =a$ at the point $B$ and $C$ respectively. Show that $\overline{AC}=\overline{AB}=a:b$

Answer 1

Let the line perpendicular to $x-axis$ be given by $x=k$

Given that this lines passes through $A$

Hence the required line is $x=a\cos \theta$

Given that this line intersects $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ at $B$

$\Rightarrow \frac{a\cos \theta }{a}\cos \theta +\frac{y}{b}\sin \theta =1$

$\Rightarrow {\cos }^2\theta +\frac{y}{b}\sin \theta =1$

$\Rightarrow \frac{y}{b}\sin \theta ={\sin }^2\theta$

$\Rightarrow y=b\sin \theta$

$\Rightarrow B=(a\cos \theta ,b\sin \theta )$

Also, Given that this line intersects $x\cos \theta +y\sin \theta =a$ at $C$

$\Rightarrow a^2\cos \theta +y\sin \theta =a$

$\Rightarrow y\sin \theta =a(1-{\cos }^2\theta )$

$\Rightarrow y=a\sin \theta$

$\Rightarrow C=(a\cos \theta ,a\sin \theta )$

Now $\Rightarrow \overline{AC}=a\sin \theta$

$\Rightarrow \overline{AB}=b\sin \theta$

Therefore, $\Rightarrow \overline{AC}:\overline{AB}=a\sin \theta :b\sin \theta$

$\Rightarrow \overline{AC}:\overline{AB}=a:b$

Hence Proved.