Question

The product of perpendicular from origin to the pair of lines $ 12{x^2} + 25xy + 12{y^2} + 11y + 2 = 0 $
a) $ \dfrac{1}{{25}} $
b) $ \dfrac{2}{{25}} $
 c) $ \dfrac{3}{{25}} $
d) $ \dfrac{4}{{25}} $

Answer 1

Hint: From the standard solution that is $ a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 $ by comparing this standard equation with the given equation we will try to find the values of unknown values after that we will try to find the product of perpendicular distance. So we have the formula as $ \dfrac{{|c|}}{{\sqrt {{{(a - b)}^2} + {{(2h)}^2}} }} $ .

Complete step-by-step answer:
Here we are asked to find the product of perpendicular from origin to the pair of lines $ 12{x^2} + 25xy + 12{y^2} + 11y + 2 = 0 $
So we have the standard equation $ a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 $ now we will find out the unknown and we will substitute that values in the formula for finding the product of perpendicular from origin to the pair of lines. So let’s find different values of unknown from the given equation:
Here we have $ a = 12,h = \dfrac{{25}}{2},b = 12,f = \dfrac{{11}}{2},c = 2 $
Hence in the above equation we will put the values so that we can find the solution.
We have the formula as $ \dfrac{{|c|}}{{\sqrt {{{(a - b)}^2} + {{(2h)}^2}} }} $ to find out the product of perpendicular from origin to the pair of lines. Now we will substitute the respective values in the given formula so as to find the final answer.
So
 $ \Rightarrow \dfrac{{|2|}}{{\sqrt {{{(12 - 12)}^2} + {{(25)}^2}} }} $
 $ = \dfrac{2}{{\sqrt {{{(25)}^2}} }} $
 $ = \dfrac{2}{{25}} $
So, the final ans is $ \dfrac{2}{{25}} $ So, the product of perpendicular from origin to the pair of lines. $ 12{x^2} + 25xy + 12{y^2} + 11y + 2 = 0 $ is $ \dfrac{2}{{25}} $ respectively. And option b is the correct option.
So, the correct answer is “Option B”.

Note: While calculating the product of perpendicular from origin to the pair of lines we have to remember the formula and while putting the values we should be careful about not misplacing the values in the formula.