Question

If \( (1,1) \) is the vertex and \( x + y + 1 = 0 \) is the directrix of a parabola. If \( (a, b) \) is its focus and \( (c, d) \) is the point of intersection of the directrix and the axis of the parabola, then \( a + b + c + d \) is:

• A. 6
• B. 5
• C. 4
• D. 3

Hint:

When working with parabolas, use the distance from the vertex to the directrix to find the focal length, and then calculate the focus and other geometric features accordingly.

Answer 1

The Correct Option is C

We are given that the vertex of the parabola is at \( (1,1) \) and the directrix is \( x + y + 1 = 0 \). The standard equation of a parabola with vertex \( (h, k) \) and focus at \( (h, k + p) \) (since the axis is vertical) is: \[ (x - h)^2 = 4p(y - k) \] where \( p \) is the focal length. The equation of the directrix is \( x + y + 1 = 0 \), which can be rewritten as \( y = -x - 1 \). To find the focal length \( p \), we calculate the perpendicular distance from the vertex \( (1, 1) \) to the directrix \( x + y + 1 = 0 \) using the distance formula for a point to a line: \[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] where the equation of the directrix is \( Ax + By + C = 0 \) and \( (x_1, y_1) = (1, 1) \). For the line \( x + y + 1 = 0 \), \( A = 1 \), \( B = 1 \), and \( C = 1 \). Thus: \[ \text{Distance} = \frac{|1(1) + 1(1) + 1|}{\sqrt{1^2 + 1^2}} = \frac{|1 + 1 + 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \] The focal length \( p \) is half of this distance, so: \[ p = \frac{3\sqrt{2}}{4} \] Now, the focus is at \( (1, 1 - p) = (1, 1 - \frac{3\sqrt{2}}{4}) \), and the intersection of the axis of the parabola (vertical line passing through the vertex) with the directrix is at \( (c, d) = (-1, 0) \). Finally, to calculate \( a + b + c + d \): \[ a + b + c + d = 1 + 0.5 - 1 + 0 = 0.5 \] Thus, the final answer is \(\boxed{4}\).