Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$.
I know how to do it with the $x$ intercepts but I can't solve this.
[Math] Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$.
algebra-precalculus
Related Solutions
Suppose the points are $A,B,C$. Then intersect the equations of perpendicular bisectors of $AB$ and $BC$. This is the center of the desired circle. (with your notation $(p,q)$)
Now calculate the distance between $(p,q)$ and $A$. Now $r$ is also found.
The question makes it clear that we haven't to check that the curve, let us name it (P), is a parabola.
1st solution: Using the form you have obtained
$$(x-4)^2+ (y-4)^2 = 2xy\tag{1}$$
The LHS being positive, we must have $xy>0$; therefore, quadrants 2 and 4 are excluded.
Besides, (P) doesn't cross the $x$ axis, because setting $y=0$ in (1) would give $16=0$. For the same reason, (P) doesn't cross the $y$ axis. Therefore (P) is entirely situated in either one of the quadrants, 1 or 3. As $(x,y)=(2,2)$ belongs to (P), only quadrant 1 is possible.
2nd solution:
3 successive facts:
exchanging $x$ and $y$ doesn't change the equation ; thus parabola (P) is symmetrical with respect to the $y=x$ straight line, thus is its axis.
point $S=(2,2)$ belongs to (P).
(P) does not intersect the $x$ axis, because, if we set $y=0$ in the equation of (P), we get the quadratic equation $x^2 - 8x +32 = 0$ that has no real roots. Thus, by symmetry, (P) does not intersect the $y$ axis.
Conclusion : (P) lies entirely in the first quadrant.
Remark: As point S is situated on the axis of symmetry of (P), it qualifies it as the apex of (P).
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Best Answer
The vertex form of the equation of a parabola is $$f(x) = a(x - h)^2 + k$$ where $(h, k)$ is the vertex of the parabola. In this case, we are given that $(h, k) = (5, 0)$. Hence, \begin{align*} f(x) & = a(x - 5)^2 + 0\\ & = a(x - 5)^2 \end{align*} Since we also know the parabola passes through the point $(7, -2)$, we can solve for $a$ because we know that $f(7) = -2$. \begin{align*} a(7 - 5)^2 & = -2\\ a(2)^2 & = -2\\ 4a & = -2\\ a & = -\frac{1}{2} \end{align*} Thus, the given parabola has equation $$f(x) = -\frac{1}{2}(x - 5)^2$$