Math, asked by DeepDutta, 1 year ago

the solving process of that question asked in the above picture

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Answered by Manikumarsingh
1
y^2=ax^2+2bx+c
diff. w.r.t y
2y=2ax.dx/dy + 2b. dx/dy
dx/dy= 2y/(2ax+2b)= y/(ax+b)
again diff. w.r.t y
d^2x/dy^2 = (ax+b).1-y.a.(dx/dy)/(ax+b)^2
= (ax+b)-y.a.y/(ax+b)/(ax+b)^2
= (ax+b)-(ax^2+2bx+c)a/(ax+b)/(ax+b)^2 = a^2x^2+2abx+b^2-a^2x^2-2abx -ac/(ax+b)^2
= b^2 - ac / (ax+b)^2
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