Question

Area of the triangle formed by the vertex, focus and one
end of latus rectum of the parabola (x + 2)2 = -12(y - 1)
is
(a) 18 (b) 36
(c) 9 (d) 12
​

Answer 1

Answer:

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Answer 2

The area of the triangle formed by focus, vertex, and end of the latus rectum of the parabola is 9 units square.

Given:

Parabola (x+2)² = -12(y-1)

To Find:

The area of the triangle is formed by the focus, vertex, and end of the latus rectum of the parabola.

Solution:

(x+2) = -12(y-1)

Put (x+2)² = X and Y = (y-1)

⇒ X² = -12Y----------(1)

This is of the form

X² = -4aY

⇒ -4aY = -12Y

⇒ a = 3

∴ Focus (0,-3)

put y=(-3) in equation (1)

X² = -12(-3)

= 36

⇒ X =√36

⇒X = ±6

(6,-3) and (-6,3 ) are the end of the latus rectum.

Vertex = (0,0)

Distance from vertex to focus is the height of the triangle

$=\sqrt{0^{2}+3^{2} } \\\\=\sqrt{9} = 3 \ units$

Distance from one latus rectum to the vertex is the Base

$= \sqrt{0^{2}+6^{2} } \\\\ =\sqrt{6^{2} } \\\\= 6\ units$

Area of the triangle = $\frac{1}{2}$ * Base * Height

=  $\frac{1}{2}$ * 6* 3

= 3 * 3

= 9 unit square.

Therefore, The area of the triangle formed by focus, vertex, and end of the latus rectum of the parabola is 9 units square.

#SPJ3

Learn More

1) If the focus of a parabola is (2,3) and its latus rectum is 8 , then find the locus of the vertex of the parabola.

Link: https://brainly.in/textbook-solutions/q-focus-parabola-2-3-latus-rectum-8

2) Find the equation of the parabola, if the focus is at (0,0) and vertex is at the intersection of the lines

Link: https://brainly.in/textbook-solutions/q-equation-parabola-focus-0-0-vertex-intersection