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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Feb 28, 2019 at 13:32
  • $\begingroup$ 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? $\endgroup$ Commented Feb 28, 2019 at 14:14

2 Answers 2


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.

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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.


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