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Which of the following line can be normal to parabola `y^2=12 x ?` `x+y-9=0` (b) `2x-y-32=0` `2x+y-36=0` (d) `3x-y-72=0`

A

x+y-9=0

B

2x-y-32=0

C

2x+y-36=0

D

3x-y-99=0

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The correct Answer is:
To determine which of the given lines can be normal to the parabola \( y^2 = 12x \), we will follow these steps: ### Step 1: Identify the parameters of the parabola The standard form of the parabola is given by \( y^2 = 4ax \). Here, we have \( 4a = 12 \), which gives us: \[ a = \frac{12}{4} = 3 \] ### Step 2: Write the equation of the normal to the parabola The equation of the normal to the parabola \( y^2 = 4ax \) at a point where the slope of the tangent is \( m \) is given by: \[ y = mx - 2am - am^3 \] Substituting \( a = 3 \) into the equation, we get: \[ y = mx - 6m - 3m^3 \] ### Step 3: Analyze each line to find its slope We will analyze each line to find its slope \( m \). 1. **Line 1: \( x + y - 9 = 0 \)** - Rearranging gives: \( y = -x + 9 \) - Slope \( m = -1 \) 2. **Line 2: \( 2x - y - 32 = 0 \)** - Rearranging gives: \( y = 2x - 32 \) - Slope \( m = 2 \) 3. **Line 3: \( 2x + y - 36 = 0 \)** - Rearranging gives: \( y = -2x + 36 \) - Slope \( m = -2 \) 4. **Line 4: \( 3x - y - 72 = 0 \)** - Rearranging gives: \( y = 3x - 72 \) - Slope \( m = 3 \) ### Step 4: Check which slopes satisfy the normal equation Now we will substitute each slope into the normal equation \( y = mx - 6m - 3m^3 \) and check if it matches the line's equation. 1. **For \( m = -1 \) (Line 1)** \[ y = -1x - 6(-1) - 3(-1)^3 = -x + 6 + 3 = -x + 9 \] This matches Line 1. 2. **For \( m = 2 \) (Line 2)** \[ y = 2x - 6(2) - 3(2)^3 = 2x - 12 - 24 = 2x - 36 \] This does not match Line 2. 3. **For \( m = -2 \) (Line 3)** \[ y = -2x - 6(-2) - 3(-2)^3 = -2x + 12 + 24 = -2x + 36 \] This matches Line 3. 4. **For \( m = 3 \) (Line 4)** \[ y = 3x - 6(3) - 3(3)^3 = 3x - 18 - 81 = 3x - 99 \] This matches Line 4. ### Conclusion The lines that can be normal to the parabola \( y^2 = 12x \) are: - Line 1: \( x + y - 9 = 0 \) - Line 3: \( 2x + y - 36 = 0 \) - Line 4: \( 3x - y - 72 = 0 \) ### Final Answer The correct options are (a), (c), and (d). ---

To determine which of the given lines can be normal to the parabola \( y^2 = 12x \), we will follow these steps: ### Step 1: Identify the parameters of the parabola The standard form of the parabola is given by \( y^2 = 4ax \). Here, we have \( 4a = 12 \), which gives us: \[ a = \frac{12}{4} = 3 \] ...
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CENGAGE ENGLISH-PARABOLA-EXERCISE (MULTIPLE CORRECT ANSWER TYPE )
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  5. If the parabola x^2=ay makes an intercept of length sqrt40 unit on the...

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  6. The equation of the directrix of the parabola with vertex at the origi...

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  7. Tangent is drawn at any point (x1, y1) other than the vertex on the pa...

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  8. The parabola y^2=4x and the circle having its center at 6, 5) intersec...

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  9. Which of the following line can be tangent to the parabola y^2=8x ? x...

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  10. If the line k^(2)(x-1)+k(y-2)+1=0 touches the parabola y^(2)-4x-4y+8=0...

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  11. The equation of a circle of radius 1 touching the circles x^2+y^2-2|x|...

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  13. The line x+ y +2=0 is a tangent to a parabola at point A, intersect t...

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  14. Which of the following line can be normal to parabola y^2=12 x ? x+y-...

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  15. A normal drawn to the parabola =4a x meets the curve again at Q such t...

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  16. A circle is drawn having centre at C (0,2) and passing through focus ...

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  17. From any point P on the parabola y^(2)=4ax, perpebdicular PN is drawn ...

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  18. Let P be the point (1,0) and Q be a point on the locus y^(2)=8x. The l...

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  19. The value(s) of a for which two curves y=ax^(2)+ax+(1)/(24)andx=ay^(2)...

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  20. From any point P on the parabola y^(2)=4ax, perpebdicular PN is drawn ...

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