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If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P is
- a)x + 2y = 42
- b)2x + 5y = 100
- c)x + 3y = 58
- d)3x + 4y = 94
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If a hyperbola passes through the point P(10, 16) and it has vertices ...
A, 0) and (-a, 0), then the equation of the hyperbola can be written in the form:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
where (h, k) is the center of the hyperbola.
We know that the hyperbola passes through the point P(10, 16), so we can substitute these values into the equation:
(10 - h)^2 / a^2 - (16 - k)^2 / b^2 = 1
We also know that the vertices are at (a, 0) and (-a, 0), so we can substitute these values into the equation:
(a - h)^2 / a^2 - (0 - k)^2 / b^2 = 1
(-a - h)^2 / a^2 - (0 - k)^2 / b^2 = 1
Since the vertices are symmetric about the y-axis, we can simplify the second equation:
(a + h)^2 / a^2 - (0 - k)^2 / b^2 = 1
Now we have a system of equations:
(10 - h)^2 / a^2 - (16 - k)^2 / b^2 = 1
(a + h)^2 / a^2 - (0 - k)^2 / b^2 = 1
To solve for the constants a, b, h, and k, we need more information or another point on the hyperbola.
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
where (h, k) is the center of the hyperbola.
We know that the hyperbola passes through the point P(10, 16), so we can substitute these values into the equation:
(10 - h)^2 / a^2 - (16 - k)^2 / b^2 = 1
We also know that the vertices are at (a, 0) and (-a, 0), so we can substitute these values into the equation:
(a - h)^2 / a^2 - (0 - k)^2 / b^2 = 1
(-a - h)^2 / a^2 - (0 - k)^2 / b^2 = 1
Since the vertices are symmetric about the y-axis, we can simplify the second equation:
(a + h)^2 / a^2 - (0 - k)^2 / b^2 = 1
Now we have a system of equations:
(10 - h)^2 / a^2 - (16 - k)^2 / b^2 = 1
(a + h)^2 / a^2 - (0 - k)^2 / b^2 = 1
To solve for the constants a, b, h, and k, we need more information or another point on the hyperbola.
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Community Answer
If a hyperbola passes through the point P(10, 16) and it has vertices ...
Let hyperbola
=
∴ (10, 16) lies on it.
⇒ b2 = 144
⇒ b = 12
∴ Equation of normal:
⇒ 2x – 20 = –5y + 80
⇒ 2x + 5y = 100
=
∴ (10, 16) lies on it.
⇒ b2 = 144
⇒ b = 12
∴ Equation of normal:
⇒ 2x – 20 = –5y + 80
⇒ 2x + 5y = 100
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If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. Can you explain this answer?
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If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. 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 If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. Can you explain this answer?.
If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. 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 If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal to it at P isa)x + 2y = 42b)2x + 5y = 100c)x + 3y = 58d)3x + 4y = 94Correct answer is option 'B'. Can you explain this answer?.
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