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Maths-

General

Easy

Question

If the system of equations ax + hy + g = 0, hx + by + f = 0 and ax² + 2hxy+ by² + 2gx + 2fy + c + t = 0 has a unique solution, then

t = 1
t = 0
$t = \frac{a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2}}{h^{2} - a b}$
$t = \frac{1}{h^{2} - a b}$

The correct answer is: $t equals fraction numerator a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent over denominator h to the power of 2 end exponent minus a b end fraction$

Clearly $a b minus h to the power of 2 end exponent not equal to 0$ [ $because$ first two equations are solvable for x, y]
Third equation can be written as
x (ax + hy + g) + y (hx + by + f) + gx + fy + c + t = 0
Since any solution of first two must be sol. of the third equation.
\ given system of equations can be written as
ax + hy + g = 0
hx + by + f = 0
gx + fy + c + t = 0
Since the above system of equations are consistent
$therefore open vertical bar table row a h g row h b f row g f cell c plus t end cell end table close vertical bar equals 0 rightwards double arrow open vertical bar table row a h g row h b f row g f c end table close vertical bar plus open vertical bar table row a h 0 row h b 0 row g f t end table close vertical bar equals 0 rightwards double arrow open parentheses a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent close parentheses plus t open parentheses a b minus h to the power of 2 end exponent close parentheses equals 0 rightwards double arrow t equals fraction numerator a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent over denominator left parenthesis h to the power of 2 end exponent minus a b right parenthesis end fraction$

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