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

Last updated date: 23rd Jul 2024
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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.

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 $\alpha \cos ec\alpha$