Question

The intercept made by line $ x\cos \alpha + y\sin \alpha = \alpha $ on y-axis is
A. $ \alpha $
B. $ \alpha \cos ec\alpha $
C. $ \alpha sec\alpha $
D. $ \alpha \sin \alpha $

Answer 1

Hint: The line equation is given in terms of trigonometric functions. So first we have to convert the line equation into intercepts form which will give us x-intercept and y-intercept. From this newly formed line equation, we can get the intercept made by line on the y-axis (y-intercept). Inverse of a cosine function is a secant function and the inverse of a sine function is a cosecant function. Use this info to further solve the problem.

Complete step-by-step answer:
We are given a line equation $ x\cos \alpha + y\sin \alpha = \alpha $ and we have to find the intercept made by this line on the y-axis (y-intercept).
Intercept form of a line equation is $ \dfrac{x}{a} + \dfrac{y}{b} = 1 $ , where a is the x-intercept and b is the y-intercept.
So first we have to convert the given line equation into this form.
 $ x\cos \alpha + y\sin \alpha = \alpha $
On dividing the above equation (both LHS and RHS) by $ \alpha $ , we get
 $ \dfrac{{x\cos \alpha + y\sin \alpha }}{\alpha } = \dfrac{\alpha }{\alpha } $
 $ \Rightarrow \dfrac{{x\cos \alpha }}{\alpha } + \dfrac{{y\sin \alpha }}{\alpha } = 1 $
The cosine function present in the above can also be written as the inverse of secant function and sine function can also be written as the inverse of cosecant function.
 $ \cos \alpha = \dfrac{1}{{sec\alpha }},\sin \alpha = \dfrac{1}{{\cos ec\alpha }} $
Substituting the above values in the equation
 $ \Rightarrow \dfrac{x}{{\alpha \left( {sec\alpha } \right)}} + \dfrac{y}{{\alpha \left( {\cos ec\alpha } \right)}} = 1 $
 $ \therefore \dfrac{x}{{\alpha sec\alpha }} + \dfrac{y}{{\alpha \cos ec\alpha }} = 1 $
Comparing the above equation with $ \dfrac{x}{a} + \dfrac{y}{b} = 1 $ , we get the $ a = \alpha sec\alpha $ and $ b = \alpha \cos ec\alpha $
This means that the y-intercept (b) is $ \alpha \cos ec\alpha $
Hence, the correct option is Option B, the intercept made by line $ x\cos \alpha + y\sin \alpha = \alpha $ on y-axis is $ \alpha \cos ec\alpha $ .
So, the correct answer is “Option B”.

Note: Another approach
Slope-intercept form of a line equation is $ y = mx + c $ , where m is the slope and c is the y-intercept.
So convert the given line equation into this slope-intercept form to find the y-intercept.
 $ x\cos \alpha + y\sin \alpha = \alpha $
 $ y\sin \alpha = \alpha - x\cos \alpha $
 $ \Rightarrow y\sin \alpha = - x\cos \alpha + \alpha $
 $ \Rightarrow y = \dfrac{{ - x\cos \alpha + \alpha }}{{\sin \alpha }} $
 $ \Rightarrow y = - \dfrac{{\cos \alpha }}{{\sin \alpha }}\left( x \right) + \dfrac{\alpha }{{\sin \alpha }} $
 $ \therefore y = - cot\alpha x + \alpha \cos ec\alpha $
As we can see the slope is $ - cot\alpha $ and the y-intercept is $ \alpha \cos ec\alpha $