Finding the vertex, axis, focus, directrix, and latus rectum of the parabola $\sqrt{x/a}+\sqrt{y/b}=1$

Find the vertex, axis, focus, directrix, and length of latus rectum of the parabola $$\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$

Try: Curve $$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$ represents a parabola.

While drawing the diagram, I am getting that the parabola touches $$x$$ axis at $$A(a,0)$$ and touches $$y$$ axis at $$(0,b)$$

Also, the tangents at points $$A$$ and $$B$$ intersect each other at $$(0,0)$$ at an angle of $$90^\circ$$.

I do not know how can I solve this problem. Could someone help me to solve it? Thanks.

• See the related question "Is $\sqrt{x/a}+\sqrt{y/b}=1$ the equation of a parabola tangent to the coordinate axes?". The question itself is not quite a duplicate, as it doesn't ask about various elements of the parabola; however, my answer identifies the focus and directrix.
– Blue
Commented Feb 28, 2019 at 13:32
• Do you know that if you select any point on the directrix and draw the tangent lines to the parabola from that point, then the tangents are perpendicular to each other? Commented Feb 28, 2019 at 14:14

2 Answers

A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point. Hence, if the tangent at vertex $$V$$ intersect the axes at $$A'=(\alpha,0)$$ and $$B'=(0,\beta)$$, then focus $$F$$ has coordinates $$(\alpha,\beta)$$.

The axis $$FV$$ is perpendicular to $$A'B'$$, hence its slope is $$\alpha/\beta$$. But the tangent at $$A$$ forms equal angles with the axis and $$FA$$, hence the slope of line $$FA$$ is $$-\alpha/\beta$$. The same goes for line $$FB$$, hence we have the equations: $${\beta-b\over\alpha}={\beta\over\alpha-a}=-{\alpha\over\beta},$$ which can be solved to find: $$\alpha={ab^2\over a^2+b^2},\quad \beta={a^2b\over a^2+b^2}.$$ The directrix is the parallel to $$A'B'$$ passing through the origin.

The given equation $$\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$ can be simplified as $$\displaystyle \sqrt{\frac{x}{a}} = 1 - \sqrt{\frac{y}{b}}$$ Squarring both sides, $$\displaystyle \frac{x}{a}= 1 - 2\sqrt{\frac{y}{b}} + \frac{y}{b}$$ $$\displaystyle \frac{x}{a}-1+ \frac{y}{b}=- 2\sqrt{\frac{y}{b}}$$ Squarring again, $$\displaystyle \left(\frac{x}{a}-1+ \frac{y}{b}\right)^2=4\frac{y}{b}$$ $$\displaystyle \left(X\right)^2=4Y$$ Where $$X=\frac{x}{a}-1+ \frac{y}{b}$$ and $$Y=\frac{y}{b}$$ Now compare with the standard form, apply the formulae and find the required properties.

Hope this helps you.