Question

The coordinate axes are rotated about the origin '$O$' in the counter clockwise direction through an angle ${60}^o$. If $a$ and $b$ are the intercepts made on the new axes by a straight line whose equation referred to the original axes is $3x+4y-5=0$ then $\frac{1}{a^2}+\frac{1}{b^2}=$

• A. $\frac{1}{25}$
• B. $\frac{1}{9}$
• C. $\frac{1}{16}$
• D. $1$

Answer 1

Answer: (D)

$1$

Let $X$ and $Y$ be the coordinates in the new coordinate system.

When axis are rotated through ${60}^{\circ }$, co-ordinates are changed to

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

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

$3x+4y=5$

In the new coordinate system, the line will be represented as

$3(X\left(\frac{1}{2}\right)-Y\left(\frac{\sqrt{3}}{2}\right))+4(X\left(\frac{\sqrt{3}}{2}\right)+Y\left(\frac{1}{2}\right))=5$

$X(3+4\sqrt{3})+Y(4-3\sqrt{3})=10$

$\Rightarrow a=\frac{10}{3+4\sqrt{3}};b=\frac{10}{4-3\sqrt{3}}$

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{{(3+4\sqrt{3})}^2+{(4-3\sqrt{3})}^2}{{\left(10\right)}^2}$

$=\frac{25+75}{100}=1$