Parabolas are drawn to touch two given rectangular axes and their foci are all at a constant distance c from the origin. Prove that the locus of the vertices of these parabolas is the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = c^{\frac{2}{3}}$ .

Open in App

Solution

We have $b = λ x^{1 / 3}$ , $a = λ y^{1 / 3}$
and $λ^{2} = \frac{(x^{2 / 3} + y^{2 / 3})^{4}}{x^{2 / 3} y^{2 / 3}}$
As co-ordinates of focus are ${\frac{a b^{2}}{a^{2} + b^{2}}, \frac{a^{2} b}{a^{2} + b^{2}}}$ , square of the distance of focus from the origin (0, 0) is
$\frac{a^{2} b^{4}}{(a^{2} + b^{2})^{2}} + \frac{a^{2} b^{2}}{(a^{2} + b^{2})^{2}} = \frac{a^{2} + b^{2}}{(a^{2} + b^{2}} = c^{2}$ (by hypothesis)

$= \frac{λ x^{2 / 3} y^{2 / 3}}{x^{2 / 3} + y^{2 / 3}}$ [by (1)] ...(2)
Eliminating $λ$ from (1) and (2), we get $c^{2 / 3} = x^{2 / 3} + y^{2 / 3}$

Suggest Corrections

Similar questions

x^{2} y^{2} = c^{2} (x^{2} + y^{2})

x^{\frac{2}{3}} y^{\frac{2}{3}} (x^{\frac{2}{3}} + y^{\frac{2}{3}}) = c^{2}

y^{2} = 4 a (x - c_{1})

x^{2} = 4 a (y - c_{2})

c_{1}

c_{2}

y^{2} = 4 a (x - l)

x 2 = 4 a (y - l^{'})

x y = 4 a^{2}

x^{2} - y^{2} - h x + k y = 0

Join BYJU'S Learning Program