The problem concerns a geometric setup involving lines and curves, requiring the determination of the value of \( a + d \). The solution is detailed below.
The given equation is: \[ x + y + 4 = 0 \] This equation defines a line with a slope of \(-1\) and a y-intercept of \(-4\). The points where this line intersects the coordinate axes are indicated in the accompanying figure. - The line crosses the x-axis at the point \( (-4, 0) \). - The line crosses the y-axis at the point \( (0, -4) \). The region whose area is to be calculated is a right triangle bounded by the x-axis, the y-axis, and the line \( x + y + 4 = 0 \).
The formula for the area of a triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] In this scenario, the base of the triangle is the segment along the x-axis from the origin \( O \) to \( A(-4, 0) \), measuring 4 units. The height is the segment along the y-axis from the origin \( O \) to \( B(0, -4) \), also measuring 4 units. Consequently, the area is computed as: \[ \text{Area} = \frac{1}{2} \times 4 \times 5 = 10 \] From this, we establish that \( a = 10 \).
The provided image illustrates the relationship between the variables \( a \) and \( d \). Based on the geometric configuration, it is understood that: \[ 6 = 4 \quad \text{therefore} \quad a + d = 14 \] Thus, the value of \( a + d \) is determined to be 14.
\[ a + d = 14 \]
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16