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Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2 – 9x2 = 1 is
- a)x2 + y2 = 9
- b)x2 + y2 = 1/9
- c)x2 + y2 = 7/144
- d)x2 + y2 = 1/16
Correct answer is option 'D'. Can you explain this answer?
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The equation of the given hyperbola is \(16y^2 - x^2 = 16\).
Let \(P(h, k)\) be any point on the hyperbola. The equation of the tangent at point \(P\) is given by:
\[16y^2 - x^2 = 16 \quad \Rightarrow \quad -x^2 = 16 - 16y^2 \quad \Rightarrow \quad x^2 = 16y^2 - 16.\]
The foot of the perpendicular drawn from the focus \((0, 0)\) to the tangent is the point \((0, k)\) (since the slope of the tangent is infinite).
The distance between the point \((h, k)\) and \((0, k)\) is \(h\). Since the foot of the perpendicular is equidistant from the two foci, the foot of the perpendicular drawn from the other focus \((0, 0)\) to the tangent is \((2h, k)\).
We need to find the locus of the points \((0, k)\) and \((2h, k)\).
From the equation of the tangent, we have:
\[x^2 = 16y^2 - 16.\]
Plugging in \(x = 0\), we get:
\[0 = 16y^2 - 16 \quad \Rightarrow \quad y^2 = 1 \quad \Rightarrow \quad y = \pm 1.\]
Therefore, the locus of the point \((0, k)\) is the set of points \((0, 1)\) and \((0, -1)\).
Plugging in \(x = 2h\), we get:
\[(2h)^2 = 16y^2 - 16 \quad \Rightarrow \quad 4h^2 = 16y^2 - 16 \quad \Rightarrow \quad h^2 = 4y^2 - 4.\]
This is the equation of a hyperbola. So, the locus of the point \((2h, k)\) is the set of points \((2h, k)\) such that \(h^2 = 4y^2 - 4\).
In conclusion, the locus of the feet of the perpendiculars drawn from either focus on a variable tangent to the hyperbola \(16y^2 - x^2 = 16\) is the set of points \((0, 1)\), \((0, -1)\), and \((2h, k)\) such that \(h^2 = 4y^2 - 4\).
Let \(P(h, k)\) be any point on the hyperbola. The equation of the tangent at point \(P\) is given by:
\[16y^2 - x^2 = 16 \quad \Rightarrow \quad -x^2 = 16 - 16y^2 \quad \Rightarrow \quad x^2 = 16y^2 - 16.\]
The foot of the perpendicular drawn from the focus \((0, 0)\) to the tangent is the point \((0, k)\) (since the slope of the tangent is infinite).
The distance between the point \((h, k)\) and \((0, k)\) is \(h\). Since the foot of the perpendicular is equidistant from the two foci, the foot of the perpendicular drawn from the other focus \((0, 0)\) to the tangent is \((2h, k)\).
We need to find the locus of the points \((0, k)\) and \((2h, k)\).
From the equation of the tangent, we have:
\[x^2 = 16y^2 - 16.\]
Plugging in \(x = 0\), we get:
\[0 = 16y^2 - 16 \quad \Rightarrow \quad y^2 = 1 \quad \Rightarrow \quad y = \pm 1.\]
Therefore, the locus of the point \((0, k)\) is the set of points \((0, 1)\) and \((0, -1)\).
Plugging in \(x = 2h\), we get:
\[(2h)^2 = 16y^2 - 16 \quad \Rightarrow \quad 4h^2 = 16y^2 - 16 \quad \Rightarrow \quad h^2 = 4y^2 - 4.\]
This is the equation of a hyperbola. So, the locus of the point \((2h, k)\) is the set of points \((2h, k)\) such that \(h^2 = 4y^2 - 4\).
In conclusion, the locus of the feet of the perpendiculars drawn from either focus on a variable tangent to the hyperbola \(16y^2 - x^2 = 16\) is the set of points \((0, 1)\), \((0, -1)\), and \((2h, k)\) such that \(h^2 = 4y^2 - 4\).
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Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. 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 Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer?.
Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. 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 Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer?.
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Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2–9x2= 1 isa)x2+ y2= 9b)x2+ y2= 1/9c)x2+ y2= 7/144d)x2+ y2= 1/16Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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