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What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?
- a)4(x – y) = 15
- b)4(x + y) = 15
- c)2(x – y) = 15
- d)2(x + y) = 15
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What will be the equation of the normal to the parabola y2 = 5x that m...
The equation of the given parabola is, y2 = 5x ……….(1)
Differentiating both sides of (1) with respect to y, we get,
2y = 5(dx/dy)
Or dx/dy = 2y/5
Take any point P((5/4)t2, (5/2)t). Then, the normal to the curve (1) at P is,
-[dx/dy]P = -(2*5t/2)/5 = -t
By the question, slope of the normal to the curve (1) at P is tan45°.
Thus, -t = 1
Or t = -1
So, the required equation of normal is,
y – 5t/2 = -t(x – 5t2/4)
Simplifying further we get,
4(x – y) = 15
Differentiating both sides of (1) with respect to y, we get,
2y = 5(dx/dy)
Or dx/dy = 2y/5
Take any point P((5/4)t2, (5/2)t). Then, the normal to the curve (1) at P is,
-[dx/dy]P = -(2*5t/2)/5 = -t
By the question, slope of the normal to the curve (1) at P is tan45°.
Thus, -t = 1
Or t = -1
So, the required equation of normal is,
y – 5t/2 = -t(x – 5t2/4)
Simplifying further we get,
4(x – y) = 15
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Community Answer
What will be the equation of the normal to the parabola y2 = 5x that m...
To find the equation of the normal to the parabola, we need to determine the slope of the tangent line first. We can differentiate the equation of the parabola to find the slope at a given point:
y^2 = 5x
Differentiating implicitly, we get:
2yy' = 5
y' = 5/(2y)
Next, we need to find the coordinates of the point on the parabola where the tangent line is to be drawn. Let's assume this point is (a, b).
Substituting these coordinates into the equation of the parabola, we have:
b^2 = 5a
Now, we substitute the slope of the tangent line (y') and the coordinates of the point (a, b) into the point-slope form of a line:
y - b = (5/(2b))(x - a)
Since the angle between the tangent line and the normal line is 45 degrees, the slopes of the two lines are negative reciprocals of each other. Therefore, the slope of the normal line is -2/5.
Substitute this slope and the coordinates of the point (a, b) into the point-slope form:
y - b = (-2/5)(x - a)
Simplifying, we have:
5(y - b) = -2(x - a)
5y - 5b = -2x + 2a
Rearranging the equation, we get:
2x + 5y = 2a + 5b
Therefore, the equation of the normal to the parabola y^2 = 5x that makes an angle of 45 degrees is 2x + 5y = 2a + 5b.
y^2 = 5x
Differentiating implicitly, we get:
2yy' = 5
y' = 5/(2y)
Next, we need to find the coordinates of the point on the parabola where the tangent line is to be drawn. Let's assume this point is (a, b).
Substituting these coordinates into the equation of the parabola, we have:
b^2 = 5a
Now, we substitute the slope of the tangent line (y') and the coordinates of the point (a, b) into the point-slope form of a line:
y - b = (5/(2b))(x - a)
Since the angle between the tangent line and the normal line is 45 degrees, the slopes of the two lines are negative reciprocals of each other. Therefore, the slope of the normal line is -2/5.
Substitute this slope and the coordinates of the point (a, b) into the point-slope form:
y - b = (-2/5)(x - a)
Simplifying, we have:
5(y - b) = -2(x - a)
5y - 5b = -2x + 2a
Rearranging the equation, we get:
2x + 5y = 2a + 5b
Therefore, the equation of the normal to the parabola y^2 = 5x that makes an angle of 45 degrees is 2x + 5y = 2a + 5b.
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What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?a)4(x – y) = 15b)4(x + y) = 15c)2(x – y) = 15d)2(x + y) = 15Correct answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?a)4(x – y) = 15b)4(x + y) = 15c)2(x – y) = 15d)2(x + y) = 15Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?a)4(x – y) = 15b)4(x + y) = 15c)2(x – y) = 15d)2(x + y) = 15Correct answer is option 'A'. Can you explain this answer?.
What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?a)4(x – y) = 15b)4(x + y) = 15c)2(x – y) = 15d)2(x + y) = 15Correct answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?a)4(x – y) = 15b)4(x + y) = 15c)2(x – y) = 15d)2(x + y) = 15Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?a)4(x – y) = 15b)4(x + y) = 15c)2(x – y) = 15d)2(x + y) = 15Correct answer is option 'A'. Can you explain this answer?.
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