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The set of real values of 'a' for which ...

The set of real values of 'a' for which at least one tangent to `y^(2)=4ax` becomes normal to the circle
`x^(2)+y^(2)-2ax-4ay+3a^(2)-0,` is

A

`[1, 2]`

B

`[sqrt2, 3]`

C

`R`

D

none of these

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The correct Answer is:
To solve the problem, we need to determine the set of real values of 'a' for which at least one tangent to the parabola \( y^2 = 4ax \) becomes normal to the given circle \( x^2 + y^2 - 2ax - 4ay + 3a^2 = 0 \). ### Step-by-Step Solution: 1. **Identify the Tangent to the Parabola**: The equation of the tangent to the parabola \( y^2 = 4ax \) can be expressed as: \[ y = mx + \frac{a}{m} \] where \( m \) is the slope of the tangent. 2. **Find the Center of the Circle**: The given circle's equation is: \[ x^2 + y^2 - 2ax - 4ay + 3a^2 = 0 \] To find the center of the circle, we can rewrite it in standard form. The center \( (h, k) \) can be determined from the coefficients: \[ h = \frac{2a}{2} = a, \quad k = \frac{4a}{2} = 2a \] Thus, the center of the circle is \( (a, 2a) \). 3. **Condition for the Tangent to be Normal to the Circle**: For the tangent line to be normal to the circle, it must pass through the center of the circle. Therefore, substituting the center \( (a, 2a) \) into the tangent line equation: \[ 2a = ma + \frac{a}{m} \] 4. **Rearranging the Equation**: Rearranging the equation gives: \[ 2a = ma + \frac{a}{m} \implies 2a - ma = \frac{a}{m} \] Dividing through by \( a \) (assuming \( a \neq 0 \)): \[ 2 - m = \frac{1}{m} \] 5. **Multiplying through by \( m \)**: Multiplying through by \( m \) yields: \[ 2m - m^2 = 1 \implies m^2 - 2m + 1 = 0 \] This simplifies to: \[ (m - 1)^2 = 0 \] Thus, \( m = 1 \). 6. **Conclusion about the Values of 'a'**: Since the equation \( (m - 1)^2 = 0 \) does not depend on 'a', it implies that for every real value of \( a \), there exists at least one tangent to the parabola that is normal to the circle. 7. **Final Result**: Therefore, the set of real values of \( a \) for which at least one tangent to the parabola becomes normal to the circle is: \[ \text{All real numbers } a \in \mathbb{R} \]

To solve the problem, we need to determine the set of real values of 'a' for which at least one tangent to the parabola \( y^2 = 4ax \) becomes normal to the given circle \( x^2 + y^2 - 2ax - 4ay + 3a^2 = 0 \). ### Step-by-Step Solution: 1. **Identify the Tangent to the Parabola**: The equation of the tangent to the parabola \( y^2 = 4ax \) can be expressed as: \[ y = mx + \frac{a}{m} ...
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OBJECTIVE RD SHARMA ENGLISH-PARABOLA-Section I - Solved Mcqs
  1. Let there be two parabolas with the same axis, focus of each being ext...

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  2. Let A and B be two points on y^(2)=4ax such that normals to the curve ...

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  3. The set of real values of 'a' for which at least one tangent to y^(2)=...

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  4. The locus of the mid-point of the line segment joining a point on the ...

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  5. The tangent and normal at the point p(18, 12) of the parabola y^(2)=8x...

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  6. Tangent and normal at any point P of the parabola y^(2)=4ax(a gt 0) me...

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  7. The points of the intersection of the curves whose parametric equation...

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  8. The locus of the midpoint of the segment joining the focus to a moving...

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  9. The radical centre of the circles drawn on the focal chords of y^(2)=4...

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  10. For each parabola y=x^(2)+px+q, meeting coordinate axes at 3-distinct ...

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  11. Let A(x(1),y(1)) and B(x(2),y(2)) be two points on the parabola y^(2) ...

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  12. Let A and B be two distinct points on the parabola y^2=4x. If the ax...

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  13. the shortest distance between the line y-x=1 and the curve x=y^(2) ...

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  14. about to only mathematics

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  15. Let PQ be a focal chord of the parabola y^(2)=4ax. The tangents to the...

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  16. Let a, r, s, t be non-zero real numbers. Let P(at^(2),2at),Q(ar^(2),2a...

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  17. Let a, r, s, t be non-zero real numbers. Let P(at^(2),2at),Q(ar^(2),2a...

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  18. Let P and Q be distinct points on the parabola y^2 = 2x such that a c...

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  19. about to only mathematics

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  20. PSQ is a focal chord of a parabola whose focus is S and vertex is A. P...

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