The values of $x$ obtained from the equation $ax^2+by^2=1$ and $ax+by=1$ will be equal if:

Question

The values of $x$ obtained from the equation $ax^2+by^2=1$ and $ax+by=1$ will be equal if: $$a). a-b=1$$ $$b). a+b=1$$ $$c). a=b$$ $$d). a=b=1$$

My Attempt: $$ax^2+by^2=1$$ $$x=\sqrt {\dfrac {1-by^2}{a}}$$ and $$ax+by=1$$ $$x=\dfrac {1-by}{a}$$ According to the Question, $$\sqrt {\dfrac {1-by^2}{a}}=\dfrac {1-by}{a}$$ On solving further, I got $$2a^2b=a^2b(y+by)$$

Answer 1

We have $$ax+by=1 \implies y =\frac{1-ax}{b}$$

Now, substituting this, we get, $$ax^2+b(\frac{1-ax}{b})^2=1 \implies abx^2+a^2x^2-2ax+1-b=0$$

Implying that the value of $x$ obtained is same from the two equations $\equiv$ the quadratic equation has only one root, we then have $$\text{ Discriminant}=0 \implies 4a^2-4(ab+a^2)(1-b)=0$$ $$\implies a^2-ab-a^2+ab^2+a^2b=0$$ $$\boxed{a+b=1}$$

Answer 2

Viewing this problem geometrically and assuming that at least one of $a$ and $b$ is non-zero, the first equation describes a (possibly degenerate) conic section, while the second is a line. For their intersections to have the same $x$-values, the line must either be vertical or tangent to the conic.

In the vertical case, we have $b=0$, so $ax^2=1$ and $ax=1$, which means that $x=a=1$. The conic in this case is degenerate, consisting of the parallel lines $x=\pm1$.

For the second case, we can work in homogeneous coordinates and write the tangency condition in terms of the dual conic’s matrix (using the adjugate to allow zeros for $a$ or $b$) and the homogeneous vector that represents the line: $$\begin{bmatrix}a&b&-1\end{bmatrix} \begin{bmatrix}-b&0&0\\0&-a&0\\0&0&ab\end{bmatrix} \begin{bmatrix}a\\b\\-1\end{bmatrix} = ab-a^2b-ab^2 = -ab(a+b-1) = 0,$$ from which we have $a=0$, $b=0$ or $a+b=1$. The condition $a=0$ allows for any pair of $x$-values, so must be rejected, and we’ve already dealt with $b=0$, leaving $a+b=1$ (which subsumes the $b=0$ case, anyway). The conic is an ellipse for $0\lt a\lt1$, a pair of parallel lines when $a=0$ or $a=1$, and a hyperbola otherwise.