Question

The coordinate axes are rotated about the origin (0,0) in counter clockwise direction through an angle of ${60^ \circ }$. lf p and q are intercepts made on new axes by a straight line whose equation referred to the original axes is $x + y = 1$, then $\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} =$ ​
A) 2
B) 4
C) 6
D) 8

Answer 1

Here first we will let X and Y be the new coordinate system. Then we will find the values of x and y in terms of ‘X’ and ‘Y’ and satisfy them in the given equation to get the values of ‘X’ and ‘Y’ which are equal to p and q. Now we will find the value of the given expression $\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}$.

Complete step-by-step answer:
Let X and Y be the new coordinate system then,

$$x = X\cos \theta - Y\sin \theta \\\ y = X\sin \theta + Y\cos \theta \\\$$

Now, since the coordinate axes are rotated about the origin (0,0) in counter clockwise direction through an angle of ${60^ \circ }$
Therefore, $\theta = {60^ \circ }$
Putting this value in above equations we get:-

$$x = X\cos {60^ \circ } - Y\sin {60^ \circ } \\\ y = X\sin {60^ \circ } + Y\cos {60^ \circ } \\\$$

As we know that,

$$\cos {60^ \circ } = \dfrac{1}{2} \\\ \sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2} \\\$$

Hence,

$$x = X\left( {\dfrac{1}{2}} \right) - Y\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\\ y = X\left( {\dfrac{{\sqrt 3 }}{2}} \right) + Y\left( {\dfrac{1}{2}} \right) \\\$$

Now satisfying these values in the given equation $x + y = 1$ we get :-

$$X\left( {\dfrac{1}{2}} \right) - Y\left( {\dfrac{{\sqrt 3 }}{2}} \right) + X\left( {\dfrac{{\sqrt 3 }}{2}} \right) + Y\left( {\dfrac{1}{2}} \right) = 1 \\\ X\left( {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}} \right) + Y\left( {\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}} \right) = 1 \\\ X\left( {\dfrac{{1 + \sqrt 3 }}{2}} \right) + Y\left( {\dfrac{{1 - \sqrt 3 }}{2}} \right) = 1 \\\$$

Further evaluating we get:-

$$X = p = \dfrac{1}{{\dfrac{{1 + \sqrt 3 }}{2}}} \\\ \Rightarrow p = \dfrac{2}{{1 + \sqrt 3 }} \\\ Y = q = \dfrac{1}{{\dfrac{{1 - \sqrt 3 }}{2}}} \\\ \Rightarrow q = \dfrac{2}{{1 - \sqrt 3 }} \\\$$

Now evaluating the value of $\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}$ we get:-

$$\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} = \dfrac{1}{{{{\left( {\dfrac{2}{{1 + \sqrt 3 }}} \right)}^2}}} + \dfrac{1}{{{{\left( {\dfrac{2}{{1 - \sqrt 3 }}} \right)}^2}}} \\\ \dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} = {\left( {\dfrac{{1 + \sqrt 3 }}{2}} \right)^2} + {\left( {\dfrac{{1 - \sqrt 3 }}{2}} \right)^2} \\\$$

Now applying the following identities:

$${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \\\ {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \\\$$

We get:-

$$\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} = \dfrac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2\left( 1 \right)\left( {\sqrt 3 } \right)}}{4} + \dfrac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2\left( 1 \right)\left( {\sqrt 3 } \right)}}{4} \\\ \dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} = \dfrac{1}{4}\left[ {1 + 3 + 2\sqrt 3 + 1 + 3 - 2\sqrt 3 } \right] \\\$$

Cancelling the terms we get:-

$$\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} = \dfrac{1}{4}\left( 8 \right) \\\ \dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}} = 2 \\\$$

Therefore, option A is correct.

Note: Transformations in the coordinate plane suggest that along the coordinate grid or plane, you can use x-axis and y-axis in order to keep track of every move. The lines also provide good assistance while drawing the polygons and flat figures. You need to concentrate on the coordinates of the objects, vertices and then join them to make the image.