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A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2 (c > 0) such that slope of BC is 2. If distance of point A from centre of xy = c2 is √10, then which of the following is/are correct for xy = c2
  • a)
    the value of c is 2
  • b)
    the value of c is 4
  • c)
    the equation of normal at point A can be y = 2x − 3√2
  • d)
    the equation of normal at point A can be y = 3x + 8√2
Correct answer is option 'A,C'. Can you explain this answer?
Verified Answer
A right angle triangle ABC, right angle at A is inscribed in hyperbola...
Let the coordinates of point A are (ct. c/t)
So. the slope of normal at A will be t2.
And normal will be parallel to BC
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Most Upvoted Answer
A right angle triangle ABC, right angle at A is inscribed in hyperbola...
We can start by finding the equation of the hypotenuse of the right triangle ABC. Let point A be (x1, y1), point B be (x2, y2), and point C be (x3, y3).

Since angle ABC is a right angle, the slope of BC multiplied by the slope of AB is -1:

(y3-y2)/(x3-x2) * (y2-y1)/(x2-x1) = -1

Simplifying this equation, we get:

(y3-y2)(x2-x1) = -(y2-y1)(x3-x2)

Expanding this equation using the coordinates of points B and C, we get:

(y3-c^2/x3 - y2)/(x3-x2) * (y2-c^2/x2 - y1)/(x2-x1) = -1

Simplifying, we get:

(y3-y2)*x2*x3 + (c^2*y2-y3*x2)*x3 + (y1-y2)*x1*x2 + (y2*x1-c^2*y1)*x2 = 0

This is the equation of the hypotenuse of the right triangle ABC. We can simplify it further by using the fact that the length of the hypotenuse is c times the square root of 2.

Using the coordinates of point A, we can write:

x1^2 - y1^2/c^2 = c^2

Solving for y1, we get:

y1 = +/- sqrt(c^4-x1^2*c^2)/c

Since we know that angle ABC is a right angle, we can use the Pythagorean theorem to find the coordinates of points B and C:

(x2-x1)^2 + (y2-y1)^2 = c^2

(x3-x1)^2 + (y3-y1)^2 = c^2

We can use the equation of the hypotenuse to eliminate either x2 or x3, and then solve for the remaining variables. Once we have the coordinates of all three points, we can check that angle ABC is indeed a right angle.
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A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2(c > 0) such that slope of BC is 2. If distance of point A from centre of xy = c2is√10, then which of the following is/are correct for xy = c2a)the value of c is 2b)the value of c is 4c)the equation of normal at point A can bey = 2x − 3√2d)the equation of normal at point A can bey = 3x + 8√2Correct answer is option 'A,C'. Can you explain this answer?
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A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2(c > 0) such that slope of BC is 2. If distance of point A from centre of xy = c2is√10, then which of the following is/are correct for xy = c2a)the value of c is 2b)the value of c is 4c)the equation of normal at point A can bey = 2x − 3√2d)the equation of normal at point A can bey = 3x + 8√2Correct answer is option 'A,C'. 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 A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2(c > 0) such that slope of BC is 2. If distance of point A from centre of xy = c2is√10, then which of the following is/are correct for xy = c2a)the value of c is 2b)the value of c is 4c)the equation of normal at point A can bey = 2x − 3√2d)the equation of normal at point A can bey = 3x + 8√2Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2(c > 0) such that slope of BC is 2. If distance of point A from centre of xy = c2is√10, then which of the following is/are correct for xy = c2a)the value of c is 2b)the value of c is 4c)the equation of normal at point A can bey = 2x − 3√2d)the equation of normal at point A can bey = 3x + 8√2Correct answer is option 'A,C'. Can you explain this answer?.
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