If the equation of the pair of straight lines | vedclass

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(A) The given pair of lines is $3x^2 - 4xy + 5y^2 = 0$. The lines perpendicular to these and passing through $(a, b)$ have the equation $5(x-a)^2 + 4(

Vedclass · Accepted Answer

$\frac{38}{5}$

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(A) The given pair of lines is $3x^2 - 4xy + 5y^2 = 0$. The lines perpendicular to these and passing through $(a, b)$ have the equation $5(x-a)^2 + 4(x-a)(y-b) + 3(y-b)^2 = 0$.Expanding this: $5(x^2 - 2ax + a^2) + 4(xy - bx - ay + ab) + 3(y^2 - 2by + b^2) = 0$.$5x^2 + 4xy + 3y^2 - (10a + 4b)x - (4a + 6b)y + (5a^2 + 4ab + 3b^2) = 0$.Comparing with $lx^2 + 2hxy + my^2 - 32x - 26y + c = 0$,we get $l=5, 2h=4 \implies h=2, m=3$.$10a + 4b = 32 \implies 5a + 2b = 16$.$4a + 6b = 26 \implies 2a + 3b = 13$.Solving these,$a=2, b=3$.Then $c = 5(2)^2 + 4(2)(3) + 3(3)^2 = 20 + 24 + 27 = 71$.Finally,$\frac{a+b+c}{l+h+m} = \frac{2+3+71}{5+2+3} = \frac{76}{10} = \frac{38}{5}$.

If the equation of the pair of straight lines intersecting at $(a, b)$ and perpendicular to the pair of lines $3x^2 - 4xy + 5y^2 = 0$ is $lx^2 + 2hxy + my^2 - 32x - 26y + c = 0$,then $\frac{a+b+c}{l+h+m} =$

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