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If `a x^2+2h x y+b y^2=1,t h e n(d^(2y))/(dx^2)`is`(h^2-a b)/((h x+b y)^2)`(b) `(a b-h^2)/((h x+b y)^2)``(h^2+a b)/((h x+b y)^2)`(d) none of theseA. `(h^(2)-ab)/((hx+by)^(3))`B. `(ab-h^(2))/(hx+by)^(2)`C. `(h^(2)+ab)/(hx+by)^(2)`D. none of these |
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Answer» `ax^(2)+2hxy+by^(2)=1` Differentiating both sides w.r.t. x, we get `2ax+2hx(dy)/(dx)+2hy+2by(dy)/(dx)=0` `"or "(dy)/(dx)=-(ax+hy)/(hx+by)` Again differentiating w.r.t. x, we get `(d^(2)y)/(dx^(2))=-[((hx+by)(a+h(dy)/(dx))-(ax+hy)(h+b(dy)/(dx)))/((hx+by)^(2))]` `=-([y(ab-h^(2))+(dy)/(dx)(h^(2)x-abx)])/((hx+by)^(2))` `=((h^(2)-ab)(y-x(dy)/(dx)))/((hx+by)^(2))` `=((h^(2)-ab))/((hx+by)^(2))[y+x(ax+hy)/(hx+by)]` `(h^(2)-ab)/((hx+by)^(3))` |
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