Home
Class 12
MATHS
y^(2)=16x is a parabola and x^(2)+y^(2)...

`y^(2)=16x` is a parabola and `x^(2)+y^(2)=16` is a circle . Then

A

circle passes through the vertex of the parabola

B

circle touches the parabola at the vertex

C

circle passes through the focus of the parabola

D

circle lies inside the parabola .

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the given equations of the parabola and the circle, and determine the relationship between them. ### Step-by-Step Solution: 1. **Identify the equations**: - The equation of the parabola is given as \( y^2 = 16x \). - The equation of the circle is given as \( x^2 + y^2 = 16 \). 2. **Convert the parabola to standard form**: - The standard form of a parabola is \( y^2 = 4ax \). - Here, we can see that \( 4a = 16 \), so \( a = 4 \). - The focus of the parabola, which is at the point \( (a, 0) \), is therefore \( (4, 0) \). 3. **Identify the circle's center and radius**: - The equation of the circle \( x^2 + y^2 = r^2 \) indicates that the center is at \( (0, 0) \) and the radius \( r \) is \( \sqrt{16} = 4 \). 4. **Find the vertex of the parabola**: - The vertex of the parabola \( y^2 = 16x \) is at the origin \( (0, 0) \). 5. **Check if the circle passes through the vertex of the parabola**: - Substitute the vertex \( (0, 0) \) into the circle's equation: \[ 0^2 + 0^2 = 16 \implies 0 \neq 16 \] - Thus, the circle does not pass through the vertex of the parabola. 6. **Check if the circle touches the parabola at the vertex**: - Since the circle does not pass through the vertex, it cannot touch the parabola at that point. 7. **Check if the circle passes through the focus of the parabola**: - The focus of the parabola is \( (4, 0) \). - Substitute \( (4, 0) \) into the circle's equation: \[ 4^2 + 0^2 = 16 \implies 16 = 16 \] - Therefore, the circle does pass through the focus of the parabola. 8. **Check if the circle is inside the parabola**: - The parabola opens to the right, and since the radius of the circle is 4 and the focus is at (4, 0), the circle cannot be entirely inside the parabola. ### Conclusion: - The circle does not pass through the vertex of the parabola. - The circle does not touch the parabola at the vertex. - The circle passes through the focus of the parabola. - The circle is not inside the parabola. Thus, the correct option is that the circle passes through the focus of the parabola.
Promotional Banner

Topper's Solved these Questions

  • PARABOLA

    MCGROW HILL PUBLICATION|Exercise SOLVED EXAMPLES LEVEL-1 (single correct answer type questions )|30 Videos
  • PARABOLA

    MCGROW HILL PUBLICATION|Exercise SOLVED EXAMPLES LEVEL-2 (single correct answer type questions )|10 Videos
  • PARABOLA

    MCGROW HILL PUBLICATION|Exercise QUESTIONS FROM PREVIOUS YEARS. B-ARCHITECTRE ENTRANCE EXAMINATION PAPERS|9 Videos
  • MATRICES

    MCGROW HILL PUBLICATION|Exercise Questions from Previous Years. B -Architecture Entrance Examination Papers|22 Videos
  • PERMUTATIONS AND COMBINATIONS

    MCGROW HILL PUBLICATION|Exercise Questions from Previous Years. B-Architecture Entrance Examination Papers |17 Videos

Similar Questions

Explore conceptually related problems

The focus of the parabola x ^(2) =-16 y is

If the common chord of the circles x^(2) + ( y -lambda)^(2) =16 and x^(2) +y^(2) =16 subtend a right angle at the origin then ' lambda' is equal to :

If the common chord of the circles x^(2)+(y-2)^(2)=16 and x^(2)+y^(2)=16 subtend a angle at the origin then lambda is equal to

The director circle of the parabola (y-2)^(2)=16(x+7) touches the circle (x-1)^(2)+(y+1)^(2)=r^(2) then r is equal to

If m is the slope of a common tangent of the parabola y^(2)=16x and the circle x^(2)+y^(2)=8 , then m^(2) is equal to

Sketch the region common to the circle x^2 +y^2=16 and the parabola x^2=6y Also find the area of the region using intergration