Question

if ax^2+2hxy+by^2=1 then (d^2y)/(dx^2)=(h^2-ab)/(hx+by)^3​

Answer 1

$\quad\quad\quad\quad\bold{\dfrac{d^{2}y}{dx^{2}}=\dfrac{h^{2}-ab}{(hx+by)^{3}}}$

Step-by-step explanation:

Given, $\mathrm{ax^{2}+2hxy+by^{2}=1}$

Differentiating both sides w. r. to x, we get

$\quad\mathrm{\dfrac{d}{dx}(ax^{2}+2hxy+by^{2})=\dfrac{d}{dx}(1)}$

$\to \mathrm{2ax+2hx\:\dfrac{dy}{dx}+2hy+2by\:\dfrac{dy}{dx}=0}$

$\to \mathrm{(hx+by)\:\dfrac{dy}{dx}=-ax-hy\quad .....(1)}$

Again, differentiating both sides w. r. to x, we get

$\quad\mathrm{\dfrac{d}{dx}\big[(hx+by)\:\dfrac{dy}{dx}\big]=\dfrac{d}{dx}(-ax-hy)}$

$\to \mathrm{(hx+by)\dfrac{d^{2}y}{dx^{2}}+(h+b\:\dfrac{dy}{dx})\:\dfrac{dy}{dx}=-a-h\:\dfrac{dy}{dx}}$

$\to \mathrm{(hx+by)\:\dfrac{d^{2}y}{dx^{2}}+h\:\dfrac{dy}{dx}+b\:(\dfrac{dy}{dx})^{2}=-a-h\:\dfrac{dy}{dx}}$

$\to \mathrm{(hx+by)\:\dfrac{d^{2}y}{dx^{2}}+h\:\dfrac{-ax-hy}{hx+by}+b\:(\dfrac{-ax-hy}{hx+by})^{2}=-a-h\:\dfrac{-ax-hy}{hx+by},\:by\:(1)}$

$\to \mathrm{(hx+by)\:\dfrac{d^{2}y}{dx^{2}}=-a-2h\:\dfrac{-ax-hy}{hx+by}-b\:(\dfrac{-ax-hy}{hx+by})^{2}}$

$\quad \mathrm{=\dfrac{-a\:(hx+by)^{2}+2h\:(ax+hy)(hx+by)-b\:(ax+hy)^{2}}{(hx+by)^{2}}}$

$\quad \mathrm{=\dfrac{-ah^{2}x^{2}-2ahxy-ab^{2}y^{2}+2ah^{2}x^{2}+2bh^{2}y^{2}+2abhxy+2h^{3}xy-a^{2}bx^{2}-2abhxy-bh^{2}y^{2}}{(hx+by)^{2}}}$

$\quad \mathrm{=\dfrac{(h^{2}-ab)\:(ax^{2}+2hxy+by^{2})}{(hx+by)^{2}}}$

$\quad \mathrm{=\dfrac{h^{2}-ab}{(hx+by)^{2}}}$

$\implies \boxed{\mathrm{\dfrac{d^{2}y}{dx^{2}}=\dfrac{h^{2}-ab}{(hx+by)^{3}}}}$

Hence, proved.

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