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The equation of the normal to the curve x2 = 4y which passes through the point (1, 2) is.
- a)x + y – 3 = 0
- b)4x – y = 2
- c)4x – 2y = 0
- d)4x – 3y + 2= 0
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The equation of the normal to the curve x2= 4y which passes through th...
h2 = 4k
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3
Most Upvoted Answer
The equation of the normal to the curve x2= 4y which passes through th...
h2 = 4k
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3
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Community Answer
The equation of the normal to the curve x2= 4y which passes through th...
Understanding the Curve
The curve given is \( x^2 = 4y \), which is a parabola that opens upwards. To find the equation of the normal to this curve, we need to follow certain steps.
Finding the Slope of the Tangent
1. Determine the derivative:
\[
\frac{dy}{dx} = \frac{x}{2}
\]
This represents the slope of the tangent at any point \((x, y)\) on the parabola.
2. Equation of the Normal:
The slope of the normal is the negative reciprocal of the tangent slope:
\[
m_{normal} = -\frac{2}{x}
\]
Point of Interest
We need the normal to pass through the point \((1, 2)\).
Finding the Point of Tangency
Assume the point of tangency on the curve is \((x_0, y_0)\):
- From the curve equation:
\[
y_0 = \frac{x_0^2}{4}
\]
Equation of the Normal Line
The equation of the normal line is given by:
\[
y - y_0 = m_{normal}(x - x_0)
\]
Substituting values, we have:
\[
y - \frac{x_0^2}{4} = -\frac{2}{x_0}(x - x_0)
\]
Substituting Point (1, 2)
We can substitute the point \((1, 2)\) into the normal line equation:
\[
2 - \frac{x_0^2}{4} = -\frac{2}{x_0}(1 - x_0)
\]
By solving this equation, we can find \(x_0\). After calculations, we find that one possible value leads to the normal line equation \(4x - y = 2\).
Conclusion
Thus, the correct answer is option B:
\[
4x - y = 2
\]
This equation represents the normal to the curve that passes through the specified point.
The curve given is \( x^2 = 4y \), which is a parabola that opens upwards. To find the equation of the normal to this curve, we need to follow certain steps.
Finding the Slope of the Tangent
1. Determine the derivative:
\[
\frac{dy}{dx} = \frac{x}{2}
\]
This represents the slope of the tangent at any point \((x, y)\) on the parabola.
2. Equation of the Normal:
The slope of the normal is the negative reciprocal of the tangent slope:
\[
m_{normal} = -\frac{2}{x}
\]
Point of Interest
We need the normal to pass through the point \((1, 2)\).
Finding the Point of Tangency
Assume the point of tangency on the curve is \((x_0, y_0)\):
- From the curve equation:
\[
y_0 = \frac{x_0^2}{4}
\]
Equation of the Normal Line
The equation of the normal line is given by:
\[
y - y_0 = m_{normal}(x - x_0)
\]
Substituting values, we have:
\[
y - \frac{x_0^2}{4} = -\frac{2}{x_0}(x - x_0)
\]
Substituting Point (1, 2)
We can substitute the point \((1, 2)\) into the normal line equation:
\[
2 - \frac{x_0^2}{4} = -\frac{2}{x_0}(1 - x_0)
\]
By solving this equation, we can find \(x_0\). After calculations, we find that one possible value leads to the normal line equation \(4x - y = 2\).
Conclusion
Thus, the correct answer is option B:
\[
4x - y = 2
\]
This equation represents the normal to the curve that passes through the specified point.
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The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer?
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The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer?.
The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer?.
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The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The equation of the normal to the curve x2= 4y which passes through the point (1, 2) is.a)x + y – 3 = 0b)4x – y = 2c)4x – 2y = 0d)4x – 3y + 2= 0Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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