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The equation of the normal to the curve parametrically represented by x=t2+3t−8 and y=2t2−2t−5
 at the point P(2,−1)  is:
  • a)
    2 x+3 y−1=0
  • b)
    6x-7y - 11 = 0
  • c)
    7x + 6y -8 = 0
  • d)
    3 x + y − 1 = 0
Correct answer is option 'C'. Can you explain this answer?
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The equation of the normal to the curve parametrically represented by ...
Given that
Now, we know that
The equation of normal to the curve y=f(x) at (x1, y1) is
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The equation of the normal to the curve parametrically represented by ...
We can start by finding the derivative of the curve with respect to t:

dx/dt = 2t + 3

To find the slope of the normal, we need to take the negative reciprocal of this derivative:

m = -1/(dx/dt) = -1/(2t + 3)

Now we need to find the point on the curve where the normal intersects. Let's call this point (x0, y0). We can find this point by plugging in a specific value of t:

x0 = t^2
y0 = 3t

Let's choose t = 1 for simplicity:

x0 = 1
y0 = 3

Now we have the slope of the normal and the point where it intersects the curve. We can use the point-slope form of a line to find the equation of the normal:

y - y0 = m(x - x0)

Plugging in the values we found, we get:

y - 3 = (-1/(2t + 3))(x - 1)

Simplifying, we get:

y = (-x/(2t + 3)) + (6t + 9)/(2t + 3)

So the equation of the normal to the curve at t = 1 is:

y = (-x/5) + 3
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The equation of the normal to the curve parametrically represented by x=t2+3t−8and y=2t2−2t−5at the point P(2,−1)is:a)2 x+3 y−1=0b)6x-7y - 11 = 0c)7x + 6y -8 = 0d)3 x + y − 1 = 0Correct answer is option 'C'. Can you explain this answer?
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The equation of the normal to the curve parametrically represented by x=t2+3t−8and y=2t2−2t−5at the point P(2,−1)is:a)2 x+3 y−1=0b)6x-7y - 11 = 0c)7x + 6y -8 = 0d)3 x + y − 1 = 0Correct answer is option 'C'. Can you explain this answer? for JEE 2024 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 parametrically represented by x=t2+3t−8and y=2t2−2t−5at the point P(2,−1)is:a)2 x+3 y−1=0b)6x-7y - 11 = 0c)7x + 6y -8 = 0d)3 x + y − 1 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the normal to the curve parametrically represented by x=t2+3t−8and y=2t2−2t−5at the point P(2,−1)is:a)2 x+3 y−1=0b)6x-7y - 11 = 0c)7x + 6y -8 = 0d)3 x + y − 1 = 0Correct answer is option 'C'. Can you explain this answer?.
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