If the system of equations ax + hy + g = 0 ¸ hx + by + f = 0 ¸ ax² + 2hxy + by² + 2gx +2fy + c + t = 0 has a unique solution and (abc + 2fgh − af² − bg² − ch²) ÷ (h² − ab) = 8 , then find t

Asked by yeeshuraj3 | 25th Jul, 2016, 05:24: PM

Expert Answer:

begin mathsize 12px style ax plus space hy space plus space straight g space equals space 0.. left parenthesis straight i right parenthesis
hx plus space by space plus space straight f space equals space 0.. left parenthesis ii right parenthesis
ax squared plus 2 hxy space plus space by squared plus 2 gx plus 2 fy plus straight c plus straight t space equals 0.............. left parenthesis iii right parenthesis

Solving space left parenthesis straight i right parenthesis space space and space left parenthesis ii right parenthesis space simultaneously space we space get comma
straight x space equals space minus fraction numerator straight h open parentheses fa minus hg close parentheses plus straight g open parentheses straight h squared minus ab close parentheses over denominator straight a open parentheses straight h squared minus ab close parentheses end fraction space space comma space space straight y space equals fraction numerator open parentheses fa minus hg close parentheses over denominator open parentheses straight h squared minus ab close parentheses end fraction

Given space that comma

fraction numerator left parenthesis abc space plus space 2 fgh space minus af squared minus bg squared minus ch squared right parenthesis over denominator open parentheses straight h squared minus ab close parentheses end fraction equals 8
rightwards double arrow negative straight c open parentheses straight h squared minus ab close parentheses plus straight f left parenthesis gh minus af right parenthesis plus straight g left parenthesis fh minus bg right parenthesis space equals space 8 open parentheses straight h squared minus ab close parentheses
rightwards double arrow straight c space equals fraction numerator straight f left parenthesis gh minus af right parenthesis plus straight g left parenthesis fh minus bg right parenthesis minus space 8 open parentheses straight h squared minus ab close parentheses over denominator open parentheses straight h squared minus ab close parentheses end fraction

Substituting space values space of space straight x space and space straight y space and space straight c space in space left parenthesis iii right parenthesis space space you space can space get space the space answer space for space straight t. end style

Answered by Vijaykumar Wani | 26th Jul, 2016, 04:56: PM