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The tangent and normal at the point P(4, 4) to the parabola, y2 = 4x intersect the x–axis at the points Q and R respectively. Then the cirucm centre of the ΔPQR is
- a)(2, 0)
- b)(2, 1)
- c)(1, 0)
- d)(1, 2)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The tangent and normal at the point P(4, 4) to the parabola, y2 = 4x i...
Eq. of tangent 2y = x + 4
∴ Q ≡ (-4,0)
Eq. of normal is y - 4 = -2 (x - 4)
⇒ y + 2x = 12
Clearly QR is diameter of the required circle.
⇒ (x + 4) (x – 6) + y2 = 0
⇒ x2 + y2 – 2x – 24 = 0
centre (1, 0)
Most Upvoted Answer
The tangent and normal at the point P(4, 4) to the parabola, y2 = 4x i...
Solution:
Given:
Point P(4, 4)
Equation of the parabola: y² = 4x
Finding the equation of the tangent at point P:
Differentiating the equation of the parabola, we get:
dy/dx = 2
At point P(4, 4), the slope of the tangent is 2.
So, the equation of the tangent at point P is y - 4 = 2(x - 4)
=> y = 2x - 4
Finding the equation of the normal at point P:
Since the slope of the normal is the negative reciprocal of the slope of the tangent, the slope of the normal is -1/2.
So, the equation of the normal at point P is y - 4 = -1/2(x - 4)
=> y = -1/2x + 6
Finding the points of intersection with the x-axis:
For the tangent:
When y = 0, x = 2
So, point Q is (2, 0)
For the normal:
When y = 0, x = 12/5
So, point R is (12/5, 0)
Calculating the center of the triangle PQR:
The midpoint of QR is the center of the triangle PQR.
Midpoint of QR = ((2 + 12/5)/2, (0 + 0)/2)
=> ((22/5)/2, 0)
=> (11/5, 0)
Therefore, the center of the triangle PQR is at (11/5, 0), which is approximately (1, 0).
Therefore, the correct answer is option 'c' (1, 0).
Given:
Point P(4, 4)
Equation of the parabola: y² = 4x
Finding the equation of the tangent at point P:
Differentiating the equation of the parabola, we get:
dy/dx = 2
At point P(4, 4), the slope of the tangent is 2.
So, the equation of the tangent at point P is y - 4 = 2(x - 4)
=> y = 2x - 4
Finding the equation of the normal at point P:
Since the slope of the normal is the negative reciprocal of the slope of the tangent, the slope of the normal is -1/2.
So, the equation of the normal at point P is y - 4 = -1/2(x - 4)
=> y = -1/2x + 6
Finding the points of intersection with the x-axis:
For the tangent:
When y = 0, x = 2
So, point Q is (2, 0)
For the normal:
When y = 0, x = 12/5
So, point R is (12/5, 0)
Calculating the center of the triangle PQR:
The midpoint of QR is the center of the triangle PQR.
Midpoint of QR = ((2 + 12/5)/2, (0 + 0)/2)
=> ((22/5)/2, 0)
=> (11/5, 0)
Therefore, the center of the triangle PQR is at (11/5, 0), which is approximately (1, 0).
Therefore, the correct answer is option 'c' (1, 0).
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The tangent and normal at the point P(4, 4) to the parabola, y2 = 4x intersect the x–axis at the points Q and R respectively. Then the cirucm centre of the ΔPQR isa)(2, 0)b)(2, 1)c)(1, 0)d)(1, 2)Correct answer is option 'C'. Can you explain this answer?
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