The equation of the pair of straight lines pa | vedclass

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(B) The pair of lines perpendicular to $3x^2-4xy+5y^2=0$ is given by $5x^2+4xy+3y^2=0$. Since the required pair of lines passes through $(2,3)$,we rep

Vedclass · Accepted Answer

$52$

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(B) The pair of lines perpendicular to $3x^2-4xy+5y^2=0$ is given by $5x^2+4xy+3y^2=0$. Since the required pair of lines passes through $(2,3)$,we replace $x$ with $(x-2)$ and $y$ with $(y-3)$: $5(x-2)^2+4(x-2)(y-3)+3(y-3)^2=0$ Expanding this: $5(x^2-4x+4)+4(xy-3x-2y+6)+3(y^2-6y+9)=0$ $5x^2-20x+20+4xy-12x-8y+24+3y^2-18y+27=0$ $5x^2+4xy+3y^2-32x-26y+71=0$ Comparing this with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get: $a=5, b=3, c=71, 2h=4$ $\Rightarrow h=2, 2g=-32$ $\Rightarrow g=-16, 2f=-26$ $\Rightarrow f=-13$. Calculating the sum: $a+b+c+f+g+h = 5+3+71-13-16+2 = 52$.

The equation of the pair of straight lines passing through the point $(2,3)$ and perpendicular to the pair of lines $3x^2-4xy+5y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+c=0$. Then $a+b+c+f+g+h=$

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