Question

The pair of straight lines perpendicular to the pair of lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ has the equation
A. $a{{x}^{2}}-2hxy+b{{y}^{2}}=0$
B. $a{{x}^{2}}+2hxy+b{{y}^{2}}n=0$
C. $a{{x}^{2}}+2hxy-b{{y}^{2}}=0$
D. $a{{y}^{2}}-2hxy+b{{x}^{2}}=0$

Answer 1

Hint: If two pairs of straight line is perpendicular to each other then the lines of one pair of straight line is perpendicular to the lines of another pairs of straight line. The slope of one set of lines is the inverse of the slopes of another set of line.

Formula Used: 3. $y=mx+c$
4. $a{{x}^{2}}+2hxy+b{{y}^{2}}$

Complete step by step solution: The given equation of the straight line is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$.
Let the lines of the given pair of straight lines are given as-
$ y={{m}_{1}}x$
$ y={{m}_{2}}x$

Where ${{m}_{1}},{{m}_{2}}$ are the slope of the lines.
Now we know that the sum of the two slopes which is equal to $\dfrac{-2h}{b}$ and multipication of the two slopes which is equal to $\dfrac{a}{b}$ . So, we can write ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ .
Now if another pair of straight lines is perpendicular to $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ then the lines are also perpendicular to the above lines.
So, the equation of the lines perpendicular to
$ y={{m}_{1}}x$
$y={{m}_{2}}x $
is given as follows-
$ y+\dfrac{1}{{{m}_{1}}}x=0 $
$y+\dfrac{1}{{{m}_{2}}}x=0 $
Thus the equation of the pair of straight lines is given as follows-

$(y+\dfrac{1}{{{m}_{1}}}x)(y+\dfrac{1}{{{m}_{2}}}x)=0 $
${{y}^{2}}+\dfrac{1}{{{m}_{1}}}xy+\dfrac{1}{{{m}_{2}}}xy+\dfrac{1}{{{m}_{1}}}.\dfrac{1}{{{m}_{2}}}{{x}^{2}}=0 $
$ {{m}_{1}}{{m}_{2}}{{y}^{2}}+xy\left( {{m}_{1}}+{{m}_{2}} \right)+{{x}^{2}}=0$

Putting the value of ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ we get-
$\dfrac{a}{b}{{y}^{2}}-\dfrac{2h}{b}x+{{x}^{2}}=0$
$a{{y}^{2}}-2hxy+b{{x}^{2}}=0$

Thus we can write that the pair of straight lines perpendicular to the pair of lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ has the equation$a{{y}^{2}}-2hxy+b{{x}^{2}}=0$.

Option ‘D’ is correct

Note: Two straight lines are perpendicular means the angle between the straight lines is ${{90}^{o}}$ . This angle can be determined by the formula $\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$.