Question

The length of perpendicular from the point $(a\cos \alpha ,a\sin \alpha )$ upon the straight line $y = x\tan \alpha + c,c > 0$ is
A.$c\cos \alpha$
B.$c{\sin ^2}\alpha$
C.$c{\sec ^2}\alpha$
D.$c{\cos ^2}\alpha$

Answer 1

Here, we will use distance formula and the general equation of line. Then compare the given equation with the standard equation, substitute in the distance formula and simplify the equations carefully.

Complete step-by-step answer:

The length of perpendicular from the point $(a\cos \alpha ,a\sin \alpha )$ and let us consider the line equation$y = x\tan \alpha + c,c > 0$.
We know the general equation of line: $ax + by + c = 0$
Convert the given equation in terms of the general equation.
$\Rightarrow x\tan \alpha - y + c = 0$
Above equation-
$\Rightarrow a = \tan \alpha \\\ \Rightarrow b = ( - 1) \\\ \Rightarrow c = c \\\$
Let us assume the points as $({x_0},{y_o})$
The distance formula d$= \dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
Place the values of “x”, “y”, “a” and “b”
$\Rightarrow$ $d = \dfrac{{\left| {a\cos \alpha \tan \alpha - a\sin \alpha + c} \right|}}{{\sqrt {{{(\tan \alpha )}^2} + {{( - 1)}^2}} }}$
Simplify the above equation-
$\Rightarrow$ $d = \dfrac{{\left| {a\cos \alpha \tan \alpha - a\sin \alpha + c} \right|}}{{\sqrt {({{\tan }^2}\alpha + 1)} }}$
We know that –${\tan ^2}\alpha + 1 = {\sec ^2}\alpha$, place it in the above equation –
$\Rightarrow$ $d = \dfrac{{\left| {a\cos \alpha \tan \alpha - a\sin \alpha + c} \right|}}{{\sqrt {{{\sec }^2}\alpha } }}$
Square and square-root cancel each other in the denominator of the above equation.
$\Rightarrow$ $d = \dfrac{{\left| {a\cos \alpha \tan \alpha - a\sin \alpha + c} \right|}}{{\sec \alpha }}$
Now, convert tangent function in terms of sine upon cosine.
$\Rightarrow$ $d = \dfrac{{\left| {a\cos \alpha \dfrac{{\sin \alpha }}{{\cos \alpha }} - a\sin \alpha + c} \right|}}{{\sec \alpha }}$
Same terms from the denominator and the numerator cancel each other.
$\Rightarrow$ $d = \dfrac{{\left| {a\sin \alpha - a\sin \alpha + c} \right|}}{{\sec \alpha }}$
Same terms with opposite signs cancel each other in the numerator of the equation.
$\Rightarrow$ $d = \dfrac{{\left| c \right|}}{{\sec \alpha }}$
It can be re-written as –
$\Rightarrow$ $d = \dfrac{c}{{\sec \alpha }}$
Also, we know that - $\dfrac{1}{{\sec \alpha }} = \cos \alpha$ substitute in the above equation –
$d = c\cos \alpha$
Hence, from the given multiple choices – the option A is the correct answer.
Note: Remember all the trigonometric formulas and co-relation among them. The correct substitution is the most important. Do simplification wisely taking considerations of all the given points and the line equation.