An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4. One vertex of the triangle is (0,1) and one altitude is contained in the y -axis. If the length of each side is k√3 units, then k is

Question

An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4. One vertex of the triangle is (0,1) and one altitude is contained in the y -axis. If the length of each side is k√3 units, then k is (A) (16/13) (B) (8/13) (C) (13/16) (D) (13/8)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-khyqouhbftuv.png" /> (x2/4)+(y2/1)=1 Equation of AB is y=√3x+1 x2+4(3 x2 + 2 √3 x + 1)=4 ⇒ 13x2+8√3x=0⇒ x=(- 8 √3/13) BC=(16 √3/13) units

Q. An equilateral triangle is inscribed in the ellipse whose equation is $x^{2} + 4 y^{2} = 4.$ One vertex of the triangle is $(0, 1)$ and one altitude is contained in the $y$ -axis. If the length of each side is $k 3$ units, then $k$ is

$\frac{16}{13}$

$\frac{8}{13}$

$\frac{13}{16}$

$\frac{13}{8}$

$\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$
Equation of $A B$ is $y = 3 x + 1$
$x^{2} + 4 (3 x^{2} + 23 x + 1) = 4$
$\Rightarrow 13 x^{2} + 83 x = 0 \Rightarrow x = \frac{- 8 3}{13}$
$BC = \frac{16 3}{13}$ units

Q. An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4. One vertex of the triangle is (0,1) and one altitude is contained in the y -axis. If the length of each side is k3​ units, then k is

1316​

138​

1613​

813​