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Find the Equation of the Parabola If the Focus is at (A, 0) and the Vertex is at (A', 0) - Mathematics

Find the equation of the parabola if the focus is at (a, 0) and the vertex is at (a', 0) 

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Solution

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let (x1, y1) be the coordinates of the point of intersection of the axis and directrix. 

 It is given that the vertex and the focus of a parabola are (a', 0) and (a, 0), respectively.

Thus, the slope of the axis of the parabola is zero.

And, the slope of the directrix cannot be defined.

Let the directrix intersect the axis at (rs). 

∴ \[\frac{r + a}{2} = a', \frac{s + 0}{2} = 0\]
\[ \Rightarrow r = 2a' - a, s = 0\] 

∴ Required equation of the directrix is \[x - 2a' + a = 0\] 

Let (xy) be any point on the parabola whose focus is S (a, 0), and the directrix is \[x - 2a' + a = 0\] 

Draw PM perpendicular to \[x - 2a' + a = 0\] 

Then, we have: 

\[SP = PM\]
\[ \Rightarrow S P^2 = P M^2 \]
\[ \Rightarrow \left( x - a \right)^2 + \left( y - 0 \right)^2 = \left( \frac{x - 2a' + a}{\sqrt{1}} \right)^2 \]
\[ \Rightarrow y^2 = \left( x - 2a' + a \right)^2 - \left( x - a \right)^2 \]
\[ \Rightarrow y^2 = x^2 + 4a '^2 + a^2 - 4a'x - 4aa' + 2ax - x^2 - a^2 + 2ax\]
\[ \Rightarrow y^2 = 4a '^2 - 4a'x - 4aa' + 4ax\]
\[ \Rightarrow y^2 = - 4\left( a' - a \right)\left( x - a' \right)\]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 25 Parabola
Exercise 25.1 | Q 3.4 | Page 24
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