If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y - 29 = 0, is

Q: Why is the vertex the midpoint of the focus and the foot of the directrix?

This is a geometric consequence of the definition $PS = PM$. At the vertex, the distance to the focus equals the distance to the directrix. Since the vertex lies on the axis, the segment $VS$ must equal the segment $VZ$.

Q: Why do we use $PS^2 = PM^2$ instead of $PS = PM$?

To avoid the square root. The distance formula involves $\sqrt{(x-x_0)^2 + (y-y_0)^2}$. Squaring both sides makes the algebra cleaner and avoids radical expressions, which are prone to calculation errors.

Q: Is there a shortcut to the coefficients without expanding the whole thing?

While there are general forms, the most reliable way in JEE is to derive the equation from the definition. It prevents errors and builds intuition. The expansion is just a test of your algebraic discipline.

Question

If the equation of the parabola, whose vertex is at $(5, 4)$ and the directrix is $3x + y - 29 = 0$, is $x^2 + ay^2 + bxy + cx + dy + k = 0$ then $a + b + c + d + k$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

A parabola is defined as the locus of all points $P(x, y)$ such that their distance from a fixed point (the focus $S$) is equal to their perpendicular distance from a fixed line (the directrix). Given the vertex $V(5, 4)$ and the directrix $3x + y - 29 = 0$, we first determine the focus $S$.

The axis of symmetry passes through the vertex $V$ and is perpendicular to the directrix. The slope of the directrix is $-3$, so the slope of the axis of symmetry is $\frac{1}{3}$. The equation of the axis is:
$$y - 4 = \frac{1}{3}(x - 5) \Rightarrow x - 3y + 7 = 0$$

To find the foot of the perpendicular $Z$ from the vertex to the directrix, we solve the system of equations:
$$3x + y = 29$$
$$x - 3y = -7$$

Solving this system yields $Z(8, 5)$. Since the vertex $V$ is the midpoint of the segment $SZ$, we use the midpoint formula:
$$\frac{x_S + 8}{2} = 5, \quad \frac{y_S + 5}{2} = 4$$

This calculation reveals the coordinates of the focus to be $S(2, 3)$.

### The Algebraic Heartbeat

We apply the definition $PS = PM$, or equivalently, $PS^2 = PM^2$. The squared distance from $P(x, y)$ to $S(2, 3)$ is:
$$PS^2 = (x - 2)^2 + (y - 3)^2$$

The squared perpendicular distance from $P(x, y)$ to the line $3x + y - 29 = 0$ is:
$$PM^2 = \frac{(3x + y - 29)^2}{3^2 + 1^2} = \frac{(3x + y - 29)^2}{10}$$

Equating these expressions, we obtain:
$$10[(x - 2)^2 + (y - 3)^2] = (3x + y - 29)^2$$

Expanding the left side:
$$10(x^2 - 4x + 4 + y^2 - 6y + 9) = 10x^2 + 10y^2 - 40x - 60y + 130$$

Expanding the right side:
$$(3x + y - 29)^2 = 9x^2 + y^2 + 841 + 6xy - 174x - 58y$$

### The Final Synthesis

Bringing all terms to one side of the equation:
$$10x^2 + 10y^2 - 40x - 60y + 130 - (9x^2 + y^2 + 6xy - 174x - 58y + 841) = 0$$

Simplifying the expression, we arrive at the final equation of the parabola:
$$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$$

Comparing this to the general form $x^2 + ay^2 + bxy + cx + dy + k = 0$, we identify the coefficients:
$a = 9, \quad b = -6, \quad c = 134, \quad d = -2, \quad k = -711$

The final sum is:
$$a + b + c + d + k = 9 - 6 + 134 - 2 - 711 = \mathbf{-576}$$

Animated Solution for Mathematics - Conic Sections: If the equation of the parabola, whose vertex is at $(5, 4)$ and the directrix is $3 x + y - 29 = 0$ , is $x^{2} + a y^{2} + b x y + c x + d y + k = 0$ then $a + b + c + d + k$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Given Data

The Axis of the Parabola

Finding the Foot of Perpendicular $Z$

Formula for Foot of Perpendicular

Substituting Values for $Z$

Coordinates of Foot $Z$

Relationship Between $V$ , $S$ , and $Z$

Calculating the Focus $S$

The Definition of a Parabola

Setting up the Equation

Expanding the Equation

Expanding the Right Side

Simplifying to General Form

Identifying the Coefficients

Calculating the Final Sum

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Algebraic Heartbeat

The Final Synthesis

Similar Questions

If the equation of the parabola with vertex $V (\frac{3}{2}, 3)$ and the directrix $x + 2 y = 0$ is $α x^{2} + β y^{2} - γ x y - 30 x - 60 y + 225 = 0,$ then $α + β + γ$ is equal to :

If the line $x - 1 = 0$ is the directrix of the parabola $y^{2} - k x + 8 = 0$ , then one of the values of $k$ is

Let $P_{1}$ be a parabola with vertex $(3, 2)$ and focus $(4, 4)$ and $P_{2}$ be its mirror image with respect to the line $x + 2 y = 6$ . Then the directrix of $P_{2}$ is $x + 2y = ____$ .

If the equation of the hyperbola with foci $(4, 2)$ and $(8, 2)$ is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to _____.

The focus of the parabola $y^{2} = 4 x + 16$ is the centre of the circle C of radius 5. If the values of $λ$ , for which C passes through the point of intersection of the lines $3 x - y = 0$ and $x + λ y = 4$ are $λ_{1}$ and $λ_{2}$ , $λ_{1} < λ_{2}$ , then $12 λ_{1} + 29 λ_{2}$ is equal to

Let $P$ be the parabola, whose focus is $(- 2, 1)$ and directrix is $2 x + y + 2 = 0$ . Then the sum of the ordinates of the points on $P$ , whose abscissa is $- 2$ , is

If vertex of a parabola is $(2, - 1)$ and the equation of its directrix is $4 x - 3 y = 21$ , then the length of its latus rectum is

The equation of the directrix of the parabola $y^{2} + 4 y + 4 x + 2 = 0$ is

The axis of a parabola is the line $y = x$ and its vertex and focus are in the first quadrant at distances $2$ and $22$ units from the origin, respectively. If the point $(1, k)$ lies on the parabola, then a possible value of $k$ is :-