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If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?
- a)Parabola
- b)Hyperbola
- c)Ellipse
- d)Circle
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
If, A normal is drawn at a point P(x, y) of a curve. It meets the x-ax...
To determine the type of curve passing through the point (0, k), we need to understand the properties of the normal drawn at a point P(x, y) on the curve.
1. Definition of a Normal:
A normal is a line that is perpendicular to the tangent at a given point on a curve.
2. Properties of the Normal:
- The slope of the normal is the negative reciprocal of the slope of the tangent at the given point.
- The product of the slopes of the tangent and the normal at a point on a curve is always -1.
3. Analyzing the Given Information:
In the given problem, a normal is drawn at point P(x, y) on the curve. It meets the x-axis at point Q. The length of PQ is constant and equal to k.
4. Evaluating the Coordinates of Point Q:
Since the normal meets the x-axis at point Q, we know that the y-coordinate of Q is 0. Therefore, the coordinates of Q are (x, 0).
5. Finding the Slope of the Normal:
The slope of the normal at point P(x, y) is the negative reciprocal of the slope of the tangent at that point. Since the normal is perpendicular to the x-axis, the slope of the tangent at P is 0. Therefore, the slope of the normal is undefined (or infinite).
6. Applying the Slope-Intercept Form of a Line:
Using the slope-intercept form of a line (y = mx + c), we can write the equation of the normal passing through point P(x, y) as y = mx + c. Since the slope (m) is undefined, the equation becomes x = c.
7. Determining the Value of c:
Substituting the coordinates of point P(x, y) into the equation x = c, we get x = x. Therefore, c = x.
8. Equation of the Normal:
The equation of the normal passing through point P(x, y) is x = c, where c = x.
9. Curve Passing through (0, k):
When the normal passes through the point (0, k), we substitute the values of x = 0 and c = 0 into the equation of the normal. This gives us x = 0, which means the curve passing through (0, k) is a vertical line.
10. Conclusion:
Hence, the curve passing through the point (0, k) is a vertical line, which can be considered as a special case of a circle with an infinite radius. Therefore, the correct answer is option 'D' - Circle.
1. Definition of a Normal:
A normal is a line that is perpendicular to the tangent at a given point on a curve.
2. Properties of the Normal:
- The slope of the normal is the negative reciprocal of the slope of the tangent at the given point.
- The product of the slopes of the tangent and the normal at a point on a curve is always -1.
3. Analyzing the Given Information:
In the given problem, a normal is drawn at point P(x, y) on the curve. It meets the x-axis at point Q. The length of PQ is constant and equal to k.
4. Evaluating the Coordinates of Point Q:
Since the normal meets the x-axis at point Q, we know that the y-coordinate of Q is 0. Therefore, the coordinates of Q are (x, 0).
5. Finding the Slope of the Normal:
The slope of the normal at point P(x, y) is the negative reciprocal of the slope of the tangent at that point. Since the normal is perpendicular to the x-axis, the slope of the tangent at P is 0. Therefore, the slope of the normal is undefined (or infinite).
6. Applying the Slope-Intercept Form of a Line:
Using the slope-intercept form of a line (y = mx + c), we can write the equation of the normal passing through point P(x, y) as y = mx + c. Since the slope (m) is undefined, the equation becomes x = c.
7. Determining the Value of c:
Substituting the coordinates of point P(x, y) into the equation x = c, we get x = x. Therefore, c = x.
8. Equation of the Normal:
The equation of the normal passing through point P(x, y) is x = c, where c = x.
9. Curve Passing through (0, k):
When the normal passes through the point (0, k), we substitute the values of x = 0 and c = 0 into the equation of the normal. This gives us x = 0, which means the curve passing through (0, k) is a vertical line.
10. Conclusion:
Hence, the curve passing through the point (0, k) is a vertical line, which can be considered as a special case of a circle with an infinite radius. Therefore, the correct answer is option 'D' - Circle.
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Community Answer
If, A normal is drawn at a point P(x, y) of a curve. It meets the x-ax...
Equation of the normal at a point P(x, y) is given by
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=>y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)
Thus, it is a circle.
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=>y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)
Thus, it is a circle.
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If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?a)Parabolab)Hyperbolac)Ellipsed)CircleCorrect answer is option 'D'. Can you explain this answer?
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