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Question: If the equation \(12{{x}^{2}}+7xy-p{{y}^{2}}-18x+qy+6=0\) represents a pair of perpendicular straigh...

If the equation 12x2+7xy−py2−18x+qy+6=012{{x}^{2}}+7xy-p{{y}^{2}}-18x+qy+6=0 represents a pair of perpendicular straight lines, then
1)p=12,q=11)p=12,q=1
2)p=1,q=122)p=1,q=12
3)p=−1,q=123)p=-1,q=12
4)p=1,q=−124)p=1,q=-12

Explanation

Solution

To solve this question we need to know about the general equation for the conic section. The general equation for the conic section is ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 for the lines to be perpendicular the sum of “a” and “b” should be zero, which means a+b=0a+b=0. The other fact the will be used to solve the problem will be abc+2fgh−af2−bg2−ch2=0abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 . Using these three fact we will find the values of the unknown.

Complete step-by-step solution:
The question ask us to find the value of the unknown if the pair of straight lines is perpendicular and the equation of the line is given as 12x2+7xy−py2−18x+qy+6=012{{x}^{2}}+7xy-p{{y}^{2}}-18x+qy+6=0 . The first step will be to write the general equation of the conic section which is ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 . We will now find the values of a,b,c,g,h,fa,b,c,g,h,f by equating the equation of the line with the general equation. On equating the two we get:
⇒a=12,h=72,b=−p,g=−182=−9,f=q2,c=6\Rightarrow a=12,h=\dfrac{7}{2},b=-p,g=\dfrac{-18}{2}=-9,f=\dfrac{q}{2},c=6
Now we will apply another important fact which says that the sum of coefficients of x2{{x}^{2}} and y2{{y}^{2}} is zero. On writing it mathematically we get:
⇒a+b=0\Rightarrow a+b=0
On substituting the values we get:
⇒12−p=0\Rightarrow 12-p=0
⇒p=12\Rightarrow p=12
To find the another unknown we will apply the other fact which says that for the straight line we the expression should be equal to zero, the expression is abc+2fgh−af2−bg2−ch2=0abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 .
We will be substituting the values in the above equation. On substitution we get:
⇒12×−p×6+2×q2×−9×72−12×(q2)2+p(−9)2−6(72)2=0\Rightarrow 12\times -p\times 6+2\times \dfrac{q}{2}\times -9\times \dfrac{7}{2}-12\times {{\left( \dfrac{q}{2} \right)}^{2}}+p{{\left( -9 \right)}^{2}}-6{{\left( \dfrac{7}{2} \right)}^{2}}=0
⇒−72p−63q2×−3q2+81p−1472=0\Rightarrow -72p-\dfrac{63q}{2}\times -3{{q}^{2}}+81p-\dfrac{147}{2}=0
⇒9p−63q2×−3q2−1472=0\Rightarrow 9p-\dfrac{63q}{2}\times -3{{q}^{2}}-\dfrac{147}{2}=0
The above expression has “p” and “q” as unknown. Since we have the value of “p” so we will substitute 1212 in place of “p”. On doing this we get:
⇒9×12−63q2×−3q2−1472=0\Rightarrow 9\times 12-\dfrac{63q}{2}\times -3{{q}^{2}}-\dfrac{147}{2}=0
⇒108−1472−3q2+632q−23=0\Rightarrow 108-\dfrac{147}{2}-3{{q}^{2}}+\dfrac{63}{2}q-23=0
⇒216−147−6q2+63q2=0\Rightarrow \dfrac{216-147-6{{q}^{2}}+63q}{2}=0
⇒69−6q2+63q2=0\Rightarrow \dfrac{69-6{{q}^{2}}+63q}{2}=0
⇒3(23−2q2+21q)2=0\Rightarrow \dfrac{3\left( 23-2{{q}^{2}}+21q \right)}{2}=0
⇒2q2−21q−23=0\Rightarrow 2{{q}^{2}}-21q-23=0
We will now factorize the above equation. On doing so we get q=1q=1 or q=−232q=\dfrac{-23}{2} .
∴\therefore If the equation 12x2+7xy−py2−18x+qy+6=012{{x}^{2}}+7xy-p{{y}^{2}}-18x+qy+6=0 represents a pair of perpendicular straight lines, then 1)p=12,q=11)p=12,q=1 .

Note: The major step that we need to do here is that we should remember the equation for the conic section. For different conic sections different conditions are applied. For example for two lines to be perpendicular the sum of coefficient of x2{{x}^{2}} and y2{{y}^{2}} is zero, for circle g2+f2−c=0{{g}^{2}}+{{f}^{2}}-c=0 and so on.