Problem 9
Question
If the focus of the parabola \((y-\beta)^{2}=4(x-\alpha)\) always lies between the lines \(x+y=1\) and \(x+y=3\), then (A) \(1<\alpha+\beta<2\) (B) \(0<\alpha+\beta<1\) (C) \(0<\alpha+\beta<2\) (D) none of these
Step-by-Step Solution
Verified Answer
The correct option is (C) \(0 < \alpha + \beta < 2\).
1Step 1: Identify the focus of the parabola
The given parabola equation is \[(y-\beta)^{2}=4(x-\alpha)\]This represents a parabola that opens horizontally, with its vertex at \((\alpha, \beta)\). The focus of such a parabola is located at \((\alpha + 1, \beta)\).
2Step 2: Consider the given conditions
The problem states that the focus, \((\alpha + 1, \beta)\), lies between the lines\[x+y=1 \quad \text{and} \quad x+y=3\]This translates to:\[1 < (\alpha + 1) + \beta < 3\]
3Step 3: Simplify the inequality
Substitute \(\alpha + 1\) into the inequality:\[1 < \alpha + 1 + \beta < 3\]Simplifying the inequality, we get:\[0 < \alpha + \beta < 2\]
4Step 4: Choose the correct option
The simplified inequality is \(0 < \alpha + \beta < 2\). Compare this with the given choices:(A) \(1 < \alpha + \beta < 2\) (B) \(0 < \alpha + \beta < 1\) (C) \(0 < \alpha + \beta < 2\) (D) None of these. The correct choice is (C) \(0 < \alpha + \beta < 2\).
Key Concepts
Parabola EquationFocus of a ParabolaInequality Solution
Parabola Equation
The equation \((y-\beta)^{2}=4(x-\alpha)\) represents a parabola. This particular equation is for a horizontal parabola. In general, a parabola can open vertically or horizontally. Each type has a distinct equation form.
For a vertically opening parabola, the general form is \((x-h)^2 = 4p(y-k)\). The vertex of this parabola is at the point \((h,k)\). For a horizontally opening parabola, like in this case, the form is \((y-k)^2 = 4p(x-h)\).
Here, the vertex is at \((h,k)\) or in our equation, \((\alpha, \beta)\). **p** dictates the direction and width of the opening. If **p** is positive, the parabola opens towards positive x or y axis; if negative, in the opposite direction. This gives us a framework to understand parabola equations and distinguish between orientations.
For a vertically opening parabola, the general form is \((x-h)^2 = 4p(y-k)\). The vertex of this parabola is at the point \((h,k)\). For a horizontally opening parabola, like in this case, the form is \((y-k)^2 = 4p(x-h)\).
Here, the vertex is at \((h,k)\) or in our equation, \((\alpha, \beta)\). **p** dictates the direction and width of the opening. If **p** is positive, the parabola opens towards positive x or y axis; if negative, in the opposite direction. This gives us a framework to understand parabola equations and distinguish between orientations.
Focus of a Parabola
The focus is a crucial part of a parabola and determines its direction. For a horizontally opening parabola like \((y-\beta)^{2}=4(x-\alpha)\), the focus is calculated as \((\alpha+1, \beta)\). The formula changes depending on whether the parabola is vertical or horizontal.
- For horizontal parabolas, the formula for the focus is given by \((h+p, k)\).
- For vertical parabolas, it is \((h, k+p)\).
Inequality Solution
Inequalities are crucial in determining boundaries of regions a point can reside. In our exercise, we see how inequalities define where the focus can lie between two lines. These lines: \(x+y=1\) and \(x+y=3\), form boundaries.
To ensure the focus \((\alpha + 1, \beta)\) fits between these two boundaries, an inequality is set up: \[ 1 < (\alpha + 1) + \beta < 3 \]Breaking this down: for the focus to be valid, the sum of its related components must be greater than 1, but less than 3.
To ensure the focus \((\alpha + 1, \beta)\) fits between these two boundaries, an inequality is set up: \[ 1 < (\alpha + 1) + \beta < 3 \]Breaking this down: for the focus to be valid, the sum of its related components must be greater than 1, but less than 3.
- First step involves simplifying the expression \((\alpha + 1) + \beta\) which becomes \(\alpha + \beta + 1\).
- Upon simplification, you achieve the inequality \(0 < \alpha + \beta < 2\).
Other exercises in this chapter
Problem 7
The centroid of the triangle formed by the feet of the normals from the point \((h, k)\) to the parabola \(y^{2}+4 a x\) \(=0,(a>0)\) lies on (A) \(x\)-axis (B)
View solution Problem 8
If from a point, the two tangents drawn to the parabola \(y^{2}=4 a x\) are normals to the parabola \(x^{2}=4 b y\), then (A) \(a^{2}>8 b^{2}\) (B) \(b^{2}>8 a^
View solution Problem 10
If the focal distance of an end of the minor axis of any ellipse (referred to its axes as the axes of \(x\) and \(y\) respectively) is \(k\) and the distance be
View solution Problem 12
If \(P(a \cos \theta, b \sin \theta)\) is a point on an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), then ' \(\theta\) ' is (A) angle of \(O P\) line
View solution