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
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
Updated On: Mar 2, 2025
- 575
- -575
- 576
- -576
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The Correct Option is D
Solution and Explanation
Given vertex is \((5, 4)\) and directrix \(3x + y – 29 = 0\)
Let foot of perpendicular of \((5, 4) \) on directrix be \((x_1, y_1)\).
\(\frac{x_1−5}{3}=\frac{y_1−4}{1}=\frac{−(−10)}{10}\)
\(∴ (x_1, y_1) = (8, 5)\)
So, focus of parabola will be \(S \)= \((2, 3)\)
Let \(P(x, y)\) be any point on parabola, then
\((x−2)^2+(y−3)^2=\frac{(3x+y−29)^2}{10}\)
\(⇒10(x^2+y^2−4x−6y+13)=9x2^+y^2+841+6xy−58y−174x\)
\(⇒x^2+9y^2−6xy+134x−2y−711=0\)
and given parabola
\(x^2+ay^2+bxy+cx+dy+k=0\)
\(∴ a = 9, b = –6, c = 134, d = –2, k = –711\)
\(∴ a + b + c + d + k = –576\)
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.