Home
Class 12
MATHS
In the parabola y^2 = 4ax, the length o...

In the parabola `y^2 = 4ax,` the length of the chord passing through the vertex and inclined to the axis at an angle

A

`4asqrt2`

B

`4a//sqrt2`

C

`2asqrt2`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the length of the chord passing through the vertex of the parabola \( y^2 = 4ax \) and inclined to the axis at an angle of \( \frac{\pi}{4} \), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Vertex and Angle**: The vertex of the parabola \( y^2 = 4ax \) is at the origin \( (0, 0) \). The angle of inclination given is \( \theta = \frac{\pi}{4} \). 2. **Parametrize the Chord**: The coordinates of a point on the chord can be expressed in terms of the length of the chord \( L \): \[ (x, y) = \left( L \cos \frac{\pi}{4}, L \sin \frac{\pi}{4} \right) \] Since \( \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we can rewrite the coordinates as: \[ (x, y) = \left( \frac{L}{\sqrt{2}}, \frac{L}{\sqrt{2}} \right) \] 3. **Substitute into the Parabola Equation**: Since point \( (x, y) \) lies on the parabola, we substitute these coordinates into the parabola equation \( y^2 = 4ax \): \[ \left( \frac{L}{\sqrt{2}} \right)^2 = 4a \left( \frac{L}{\sqrt{2}} \right) \] 4. **Simplify the Equation**: This gives us: \[ \frac{L^2}{2} = 4a \cdot \frac{L}{\sqrt{2}} \] Multiplying both sides by \( 2\sqrt{2} \) to eliminate the fractions: \[ L^2 = 8aL \] 5. **Rearranging the Equation**: Rearranging gives: \[ L^2 - 8aL = 0 \] Factoring out \( L \): \[ L(L - 8a) = 0 \] 6. **Finding the Length of the Chord**: This gives us two solutions: \( L = 0 \) (which corresponds to the vertex) or \( L = 8a \). Thus, the length of the chord is: \[ L = 8a \] 7. **Final Length Calculation**: Since the chord is inclined at an angle of \( \frac{\pi}{4} \), we need to account for the orientation. The effective length of the chord is: \[ L = 8a \cdot \frac{1}{\sqrt{2}} = 4a\sqrt{2} \] ### Conclusion: The length of the chord passing through the vertex and inclined to the axis at an angle of \( \frac{\pi}{4} \) is: \[ \boxed{4a\sqrt{2}} \]
Promotional Banner

Topper's Solved these Questions

  • THE PARABOLA

    ML KHANNA|Exercise Problem Set (1) (TRUE AND FALSE)|2 Videos
  • THE PARABOLA

    ML KHANNA|Exercise Problem Set (1) (FILL IN THE BLANKS)|2 Videos
  • THE HYPERBOLA

    ML KHANNA|Exercise SELF ASSESSMENT TEST |4 Videos
  • THEORY OF QUADRATIC EQUATIONS

    ML KHANNA|Exercise Self Assessment Test|27 Videos

Similar Questions

Explore conceptually related problems

In the parabola y^(2)=4ax, the length o the chord pasing through the verte and inclined toi the axis at pi/4 is 4sqrt(2)a b.2sqrt(2)a c.sqrt(2)a d. none of these

Write the length of het chord of the parabola y^(2)=4ax which passes through the vertex and in inclined to the axis at (pi)/(4)

IF (a,b) is the mid point of chord passing through the vertex of the parabola y^2=4x , then

A set of parallel chords of the parabola y^(2)=4ax have their midpoint on any straight line through the vertex any straight line through the focus a straight line parallel to the axis another parabola

The length of the chord of the parabola y^(2) = 12x passing through the vertex and making an angle of 60^(@) with the axis of x is