The coordinates of the vertex of the parabola
\[
y = 2x^2 - 12x + 26
\]
are
Show Hint
- \( (6,13) \)
- \( (3,-8) \)
- \( (3,8) \)
- \( (6,-13) \)
- \( (3,11) \)
The Correct Option is C
Solution and Explanation
The standard form of a quadratic equation is: \[ y = ax^2 + bx + c \] where \( a = 2 \), \( b = -12 \), and \( c = 26 \). The formula for the vertex is: \[ x = -\frac{b}{2a} \] \[ x = -\frac{-12}{2(2)} = \frac{12}{4} = 3 \] Now, substituting \( x = 3 \) in the given equation: \[ y = 2(3)^2 - 12(3) + 26 \] \[ y = 2(9) - 36 + 26 = 18 - 36 + 26 = 8 \] Thus, the vertex is \( (3,8) \).
Final Answer: \[ \boxed{(3,8)} \]
Top Questions on Parabola
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) 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
- Let the point $ P $ of the focal chord $ PQ $ of the parabola $ y^2 = 16x $ be $ (1, -4) $. If the focus of the parabola divides the chord $ PQ $ in the ratio $ m : n $, gcd($m, n$) = 1, then $ m^2 + n^2 $ is equal to:
Let the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
- Let P be the parabola, whose focus is $ (-2, 1) $ and directrix is $ 2x + y + 2 = 0 $. Then the sum of the ordinates of the points on P, whose abscissa is -2, is:
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