Name: Conic Sections: The Parabola
Uploaded: 2019-07-19
Duration: 9 min 43 s
Description: This video defines a parabola and explains how to graph a parabola in standard form, which will help you to solve questions in exam. Watch and learn.

Question

Show that the locus of the centroids of equilateral triangles inscribed in the parabola

y^{2} = 4 a x

is the parabola

9 y^{2} - 4 x a + 32 a^{2} = 0

.

Accepted Answer

Given parabola is $y^{2} = 4 a x$

Let $A (x_{1}, y_{1}) & B (x_{2}, y_{2}) & C (x_{3}, y_{3})$ be the vertices of an equilateral triangle inscribed in the parabola.

$∵$ Median $C D$ is also the altitude on $A B$ where $D$ is mid point of $A B$

i.e. Co-ordinate of $D \equiv (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

$∴ Slope of C D \times Slope of A B = - 1$

$\Rightarrow \frac{\frac{y_{1} + y_{2}}{2} - y_{3}}{\frac{x_{1} + x_{2}}{2} - x_{3}} \times \frac{y_{1} - y_{2}}{x_{1} - z_{2}} = - 1$

$\Rightarrow (x_{1} - x_{2}) (x_{1} + x_{2} - 2 x_{3}) + (y_{1} - y_{2}) (y_{1} + y_{2} - 2 y_{3}) = 0 . . . . (1)$

Also $(x_{1}, y_{1}) \equiv (\frac{y_{1}^{2}}{4 a}, y_{1}), (x_{2} y_{2}) \equiv (\frac{y_{2}^{2}}{4 a}, y_{2}) & (x_{3}, y_{3}) \equiv (\frac{y_{3}^{2}}{4 a}, y_{3})$ as these vertices lie on the parabola.

$∴$ Equation $(1)$ becomes

$(\frac{y_{1}^{2}}{4 a} - \frac{y_{2}^{2}}{4 a}) (\frac{y_{1}^{2}}{4 a} + \frac{y_{2}^{2}}{4 a} - 2 \frac{y_{3}^{2}}{4 a}) + (y_{1} - y_{2}) (y_{1} + y_{2} - 2 y_{3}) = 0$

$\Rightarrow (y_{1} - y_{2}) (y_{1} + y_{2}) (y_{1}^{2} + y_{2}^{2} - 2 y_{3}^{2}) + 16 a^{2} (y_{1} - y_{2}) (y_{1} + y_{2} - 2 y_{3}) = 0$

$∴ (y_{1} + y_{2}) (y_{1}^{2} + y_{2}^{2} - 2 y_{3}^{2}) + 16 a^{2} (y_{1} + y_{2} - 2 y_{3}) = 0 . . . . (2)$

$[∵ y_{1} \neq y_{2}]$

Similarly we can also write

$(y_{2} + y_{3}) (y_{2}^{2} + y_{3}^{2} - 2 y_{1}^{2}) + 16 a^{2} (y_{2} + y_{3} - 2 y_{1}) = 0 . . . . (3)$

Now equation $(2)$ $-$ equation $(3)$ , we get

$y_{1}^{3} - y_{3}^{3} + y_{2}^{2} (y_{1} - y_{3}) + y_{2} (y_{1}^{2} - y_{3}^{2}) + 2 y_{1} y_{3} (y_{1} - y_{3}) + 48 a^{2} (y_{1} - y_{3}) = 0$

$\Rightarrow (y_{1} - y_{3}) (y_{1}^{2} + y_{3}^{2} + y_{1} y_{3}) + y_{2}^{2} (y_{1} - y_{3}) + 3 y_{2} (y_{1} - y_{3}) (y_{1} + y_{3}) + 2 y_{1} y_{3} (y_{1} - y_{3}) + 48 a^{2} (y_{1} - y_{3}) = 0$

$\Rightarrow y_{1}^{2} + y_{3}^{2} + y_{1} y_{3} + y_{2}^{2} + 3 y_{2} (y_{1} + y_{3}) + 2 y_{1} y_{3} + 48 a^{2} = 0$

$\Rightarrow y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + 3 (y_{1} y_{2} + y_{2} y_{3} + y_{3} y_{1}) + 48 a^{2} = 0$

Also ${(y_{1} + y_{2} + y_{3})}^{2} = y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + 2 (y_{1} y_{2} + y_{2} y_{3} + y_{3} y_{1})$

$∴$ Equation becomes

$y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + \frac{3}{2} [{(y_{1} + y_{2} + y_{3})}^{2} - (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})] + 48 a^{2} = 0$

$\Rightarrow \frac{- 1}{2} (y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) + \frac{3}{2} {(y_{1} + y_{2} + y_{3})}^{2} + 48 a^{2} = 0 . . . . (4)$

Let $(h, k)$ be the co-ordinates of centroid.

$∴ h = \frac{\frac{y_{1}^{2}}{4 a} + \frac{y_{2}^{2}}{4 a} + \frac{y_{3}^{2}}{4 a}}{3} & k = \frac{y_{1} + y_{2} + y_{3}}{3}$

$∴ y_{1}^{2} + y_{2}^{2} + y_{3}^{2} = 12 a h & y_{1} + y_{2} + y_{3} = 3 k$

Substituting value of $(y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) & (y_{1} + y_{2} + y_{3}) in (4)$ , we get

$\frac{- 1}{2} \times 12 a h + \frac{3}{2} {(3 k)}^{2} + 48 a^{2} = 0$

$\Rightarrow \frac{27}{2} k^{2} - 6 a h + 48 a^{2} = 0$

$\Rightarrow 9 k^{2} - 4 a h + 32 a^{2} = 0$

Replace $(h, k)$ by $(x, y)$ we get,

$9 y^{2} - 4 a x + 32 a^{2} = 0$

Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^{2} = 4 a x$ is the parabola $9 y^{2} - 4 x a + 32 a^{2} = 0$ .

Important Questions on Parabola

Learn and Prepare for any exam you want !

Show that the locus of the centroids of equilateral triangles inscribed in the parabola y2=4ax is the parabola 9y2-4xa+32a2=0.

Important Questions on Parabola

Learn and Prepare for any exam you want !

Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^{2} = 4 a x$ is the parabola $9 y^{2} - 4 x a + 32 a^{2} = 0$ .