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The equation of the directrix of the par...

The equation of the directrix of the parabola with vertex at the origin and having the axis along the x-axis and a common tangent of slope 2 with the circle `x^2+y^2=5` is (are) `x=10` (b) `x=20` `x=-10` (d) `x=-20`

A

x=10

B

x=20

C

x=-10

D

x=-20

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The correct Answer is:
To solve the problem, we need to find the equation of the directrix of a parabola with a vertex at the origin, axis along the x-axis, and a common tangent of slope 2 with the circle \(x^2 + y^2 = 5\). ### Step-by-Step Solution: 1. **Identify the Parabola's Equation**: The parabola with vertex at the origin and axis along the x-axis can be represented as: \[ y^2 = 4ax \] where \(a\) is the distance from the vertex to the focus. 2. **Equation of the Circle**: The given circle is: \[ x^2 + y^2 = 5 \] This means the radius \(r\) of the circle is \(\sqrt{5}\). 3. **Equation of the Tangent to the Circle**: The general equation of the tangent to the circle \(x^2 + y^2 = r^2\) is given by: \[ y = mx \pm \sqrt{r^2(1 + m^2)} \] Here, the slope \(m = 2\) and \(r^2 = 5\). Thus, the equation becomes: \[ y = 2x \pm \sqrt{5(1 + 2^2)} = 2x \pm \sqrt{5 \cdot 5} = 2x \pm 5 \] Therefore, the tangents to the circle are: \[ y = 2x + 5 \quad \text{and} \quad y = 2x - 5 \] 4. **Equation of the Tangent to the Parabola**: The equation of the tangent to the parabola \(y^2 = 4ax\) is given by: \[ y = mx + \frac{a}{m} \] Substituting \(m = 2\): \[ y = 2x + \frac{a}{2} \] 5. **Setting the Tangents Equal**: Since the tangent is common to both the circle and the parabola, we can equate the two expressions for \(y\): \[ 2x + \frac{a}{2} = 2x + 5 \] This simplifies to: \[ \frac{a}{2} = 5 \implies a = 10 \] 6. **Finding the Directrix**: The directrix of the parabola \(y^2 = 4ax\) is given by the equation: \[ x = -a \] Substituting \(a = 10\): \[ x = -10 \] 7. **Considering the Other Parabola**: Since the parabola can also open to the left, we can also consider: \[ x = a \implies x = 10 \] ### Final Answer: Thus, the equations of the directrix for the two possible parabolas are: - \(x = -10\) - \(x = 10\) The correct options from the given choices are: - (c) \(x = -10\) - (a) \(x = 10\)

To solve the problem, we need to find the equation of the directrix of a parabola with a vertex at the origin, axis along the x-axis, and a common tangent of slope 2 with the circle \(x^2 + y^2 = 5\). ### Step-by-Step Solution: 1. **Identify the Parabola's Equation**: The parabola with vertex at the origin and axis along the x-axis can be represented as: \[ y^2 = 4ax ...
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CENGAGE ENGLISH-PARABOLA-EXERCISE (MULTIPLE CORRECT ANSWER TYPE )
  1. The locus of the midpoint of the focal distance of a variable point ...

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  2. A square has one vertex at the vertex of the parabola y^2=4a x and the...

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  3. If two distinct chords of a parabola y^2=4ax , passing through (a,2a) ...

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  4. about to only mathematics

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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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  12. about to only mathematics

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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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