TIFR CSE 2014 | Part A | Question: 13

Question

Let $L$ be a line on the two dimensional plane. $L'$s intercepts with the $X$ and $Y$ axes are respectively $a$ and $b$. After rotating the co-ordinate system (and leaving $L$ untouched), the new intercepts are $a'$ and $b'$ respectively. Which of the following is TRUE?

• A. $\frac{1}{a}+\frac{1}{b}=\frac{1}{a}+\frac{1}{b}$.
• B. $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{a'^{2}}+\frac{1}{b'^{2}}$.
• C. $\frac{b}{a^{2}}+\frac{a}{b^{2}}=\frac{b'}{a'^{2}}+\frac{a}{b'^{2}}$.
• D. $\frac{b}{a}+\frac{a}{b}=\frac{b'}{a'}+\frac{a'}{b'}$.
• E. None of the above.

Comments

if we rotate coordinates by θ , then our a' and b' would be like

a'=acosθ+bsinθ

b'=-asinθ+bcosθ

as i'm able to see , no options is matching after placing values.

https://en.wikipedia.org/wiki/Rotation_of_axes

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No simplification?

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I think answer should be (b). I have added the answer.

Answer 1

If $X-$intercept is $a$ and $Y-$intercept is $b$ then equation of line on $X-Y$ plane is

$$\frac{x}{a} + \frac{y}{b} = 1 \quad \to (1)$$

We can prove it by considering two points on the line, $(a,0)$ as $(x_1,y_1)$ and $(0,b)$ as $(x_2,y_2)$

So, equation of line will be : $y - y_1 = \left ( \frac{y_2-y_1}{x_2-x_1} \right )(x-x_1)$

$\Rightarrow (y - 0) = \left ( \frac{b-0}{0-a} \right )(x-a)$

$\Rightarrow -ay = bx - ab$

On dividing by $ab$ and rearranging,

$\Rightarrow$ $\frac{x}{a} + \frac{y}{b} = 1$

Now, if a point $(x,y)$ lies on the $X-Y$ plane and if we rotate the co-ordinate system by $\theta$ degree then this point becomes $(x',y')$ on new co-ordinate system and relationship is given by :

$\begin{bmatrix} \cos\theta & \sin\theta \\  - \sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} x'\\y' \end{bmatrix}$

So,

• $x' = x\cos\theta +y\sin\theta$ and
• $y'= - x\sin\theta + y\cos\theta$

Now, on the new co-ordinate system, equation on line will be :

$\frac{x'}{a'} + \frac{y'}{b'} = 1$

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

$\Rightarrow x\left ( \frac{\cos\theta}{a'} - \frac{\sin\theta}{b'} \right ) + y\left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right ) = 1$

On comparing with equation $(1),$

$\left ( \frac{\cos\theta}{a'} + \frac{-\sin\theta}{b'} \right ) = \frac{1}{a}$ and

$\left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right ) = \frac{1}{b}$

Now, on squaring both sides and adding above $2$ equations :-

$\left ( \frac{1}{a^2} + \frac{1}{b^2} \right ) = \left ( \frac{\cos\theta}{a'} + \frac{-\sin\theta}{b'} \right )^{2} + \left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right )^2$

$ \Rightarrow\left ( \frac{1}{a^2} + \frac{1}{b^2} \right ) = $ $\frac{\cos^{2}\theta}{a'^2} + \frac{\sin^2\theta}{b'^2} -  \frac{2\sin\theta \cos\theta}{a'b'} + \frac{\sin^2\theta}{a'^2} + \frac{\cos^2\theta}{b'^2} + \frac{2\sin\theta \cos\theta}{a'b'}$

$ \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{a'^2} + \frac{1}{b'^2}$

So, Answer is (b)

Comments

@ankitgupta.1729

It is general equation

right?

I want to why u changed it?

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@srestha mam, Please see all the comments and links.

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Good answer, 
In exam however, if comparing doesn't strike, we can also find $a$ in terms of $a'$ & $b'$by substituting the value of $(a,0)$ into the line $\frac{xcos\theta - ysin\theta}{a'}+\frac{xsin\theta + ycos\theta}{b'} = 1$

Also, we can find $b$ in terms of $a'$ & $b'$by substituting the value of $(0,b)$ into the line.

One last clarification, your rotation matrix transforms coordinates clockwise. Also, the rotation doesn't matter clockwise or anticlockwise, as the resulting property still holds.