What curve passes through the intercepts of two given lines?

Key takeaway: Once two lines cut the axes, their intercept points give you four fixed points. The trick is to recognize the family of curves that can pass through them all.

Step 1: Find the four points

For the line

$3x+4y=24,$

the intercepts are:

• when $y=0$, $x=8$ so one point is $(8,0)$
• when $x=0$, $y=6$ so one point is $(0,6)$

For the line

$4x+3y=24,$

the intercepts are:

• when $y=0$, $x=6$ so one point is $(6,0)$
• when $x=0$, $y=8$ so one point is $(0,8)$

So the four points are on the axes.

Step 2: Form the general curve

A family of curves through all four points can be written as

$(3x+4y-24)(4x+3y-24)+\lambda xy=0.$

This works because each of the four intercept points makes one of the factors zero, and the $xy$ term also disappears on either axis.

Step 3: Identify the type

When $\lambda$ changes, the equation represents different conics. Depending on the value of $\lambda$, the curve can be:

• a circle
• a parabola
• an ellipse
• a hyperbola

So the best multiple-choice answer is that the points lie on all of these possible curves depending on $\lambda$.

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Pitfalls the pros know 👇 A common slip is to assume there is one fixed conic without checking the parameter. Here the equation contains $\lambda$, so the shape is not unique. Another mistake is forgetting that intercept points lie on the axes, which is exactly why the $xy$ term is useful.

What if the problem changes? If $\lambda$ were chosen to make the quadratic symmetric in a particular way, the same framework could yield a circle or ellipse after simplification. If $\lambda$ changes sign, the curve can move from closed to open behavior, which is how an ellipse-like form can shift toward a hyperbola-like one.

Tags: intercepts, conic sections, parameterized family