To-make-the-ellipse-4-2-2-0-lie-tangent-to-the-axis-at-the-origin-and-pass-through-the-point-1-2-what-values-of-the-constants-and-should-have-

Question Number 135747 by liberty last updated on 15/Mar/21

$$$$To make the ellipse 4𝑥^2 + 𝑦^2 + 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 lie tangent to the 𝑥 axis at the origin and pass through the point (−1,2), what values of the constants 𝑎, 𝑏, and 𝑐 should have?

Answered by EDWIN88 last updated on 15/Mar/21

$$\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(-1,2)$$$$\Rightarrow 4+4-\mathrm{a}+2\mathrm{b}+\mathrm{c}=0,-\mathrm{a}+2\mathrm{b}+\mathrm{c}=-8\left(\mathrm{i}\right)$$$$\mathrm{tangent}\mathrm{line}\mathrm{at}\mathrm{origin}\Rightarrow \mathrm{y}=0$$$$\Rightarrow \frac{\mathrm{a}}{2}(0+\mathrm{x})+\frac{\mathrm{b}}{2}(0+\mathrm{y})+\mathrm{c}=0$$$$\Rightarrow \mathrm{x}+\mathrm{y}+2\mathrm{c}=0;\mathrm{y}=-\mathrm{x}-2\mathrm{c}\mathrm{we}\mathrm{get}\{\begin{array}{l}\mathrm{a}=0\\ \mathrm{c}=0\end{array}$$$$\mathrm{equation}\left(\mathrm{i}\right)\mathrm{become}2\mathrm{b}=-8,\mathrm{b}=-4$$$$\mathrm{then}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{ellipse}\mathrm{is}$$$$\to {4\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}=0$$

Answered by mr W last updated on 15/Mar/21

$$4x^2+y^2+ax+by+c=0$$$$passingthrough(0,0):$$$$4\times 0^2+0^2+a\times 0+b\times 0+c=0$$$$\Rightarrow c=0$$$$8x+2y\frac{dy}{dx}+a+b\frac{dy}{dx}=0$$$$tangentxaxisat(0,0):$$$$8\times 0+2\times 0\times 0+a+b\times 0=0$$$$\Rightarrow a=0$$$$4x^2+y^2+by=0$$$$passingthrough(-1,2):$$$$4\times {(-1)}^2+2^2+b\times 2=0$$$$\Rightarrow b=-4$$$$$$$$\Rightarrow eqn.4x^2+y^2-4y=0$$

Commented by mr W last updated on 15/Mar/21

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