The locus of the trisection point of any arbitrary double ordinate of the parabola x2=4y, is

Question

The locus of the trisection point of any arbitrary double ordinate of the parabola x2=4y, is (A) 9x2=y (B) 3x2=2y (C) 9x2=4y (D) 9x2=2y

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-1p3ibdwqlmrwz7kz.jpg" /> Let A≡ (2 t , t2),B≡ (- 2 t , t2) be the extremities on the double ordinate AB. Let C(h , k) be it&rsquo;s trisection point, then 6h=4t and k=t2 ⇒ t=(3 h/2),t2=k⇒ k=(9 h2/4) Thus, locus of C is 9x2=4y

Q. The locus of the trisection point of any arbitrary double ordinate of the parabola $x^{2} = 4 y,$ is

$9 x^{2} = y$

$3 x^{2} = 2 y$

$9 x^{2} = 4 y$

$9 x^{2} = 2 y$

Let $A \equiv (2 t, t^{2}), B \equiv (- 2 t, t^{2})$ be the extremities on the double ordinate $A B .$
Let $C (h, k)$ be it’s trisection point, then
$6 h = 4 t$ and $k = t^{2}$
$\Rightarrow t = \frac{3 h}{2}, t^{2} = k \Rightarrow k = \frac{9 h ^{2}}{4}$
Thus, locus of $C$ is $9 x^{2} = 4 y$

Q. The locus of the trisection point of any arbitrary double ordinate of the parabola x2=4y, is

9x2=y

3x2=2y

9x2=4y

9x2=2y