Identify the following conic and hence reduce it in its standard form:
5x2-24xy-5y2+4x+58y-59=0
5x2−24xy−5y2+4x+58y−59=05x^2-24xy-5y^2+4x+58y-59=05x2−24xy−5y2+4x+58y−59=0
Compare with
ax2+bxy+cy2+dx+ey+f=0a=5;b=−24;c=−5;d=4;e=58;f=−59Nowb2−4ac(−24)2−4∗5∗(−5)=676b2−4ac>0Hence a hyperbolaThe equation isx2a2−y2b2=1x252−y2(−24)2=1x225−y2576=1ax^2+bxy+cy^2+dx+ey+f=0\\ a=5; b = -24; c=-5; d=4; e= 58; f=-59\\ Now \\ b^2-4ac\\ (-24)^2-4*5*(-5)= 676\\ b^2-4ac > 0 \\ Hence \space \space a \space hyperbola \\ The \space equation \space is \\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{5^2}-\frac{y^2}{(-24)^2}=1\\ \frac{x^2}{25}-\frac{y^2}{576}=1\\ax2+bxy+cy2+dx+ey+f=0a=5;b=−24;c=−5;d=4;e=58;f=−59Nowb2−4ac(−24)2−4∗5∗(−5)=676b2−4ac>0Hence a hyperbolaThe equation isa2x2−b2y2=152x2−(−24)2y2=125x2−576y2=1
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