The area of the triangle formed by the coordinate axes and tangent at vertex to the parabola whose focus is $(3,4)$ and tangents at $x=0$ and $y=0$ is?

I know how to do this, assume an equation for parabola with axis as $y=4/3 x$ and do all the procedure. But, this isn't how it is really to be solved. There's a sleek way which I cannot find.

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    $\begingroup$ This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $\left(h, k+ \frac{1}{4a}\right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ \frac{1}{4a}= 4$ so $k= 4-\frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- \frac{1}{4a}$, for any a, has focus (3, 4). $\endgroup$
    – user247327
    Jan 3, 2019 at 12:19
  • $\begingroup$ @user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola. $\endgroup$
    – Iceberry
    Jan 3, 2019 at 12:21
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    $\begingroup$ How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis? $\endgroup$
    – amd
    Jan 3, 2019 at 21:46

1 Answer 1


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 the tangent at the vertex intersects the axes at $(3,0)$ and $(0,4)$.

enter image description here


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