Question:
The equation of the director circle of the hyperbola $\frac {x^{2}} {16}-\frac{y^{2}} {4}=1$ is given by .............
The equation of the director circle of the hyperbola $\frac {x^{2}} {16}-\frac{y^{2}} {4}=1$ is given by .............
Updated On: Apr 18, 2024
- $x^2 + y^2 = 16$
- $x^2 + y^2 = 4$
- $x^2 + y^2 = 20$
- $x^2 + y^2 = 12$
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The Correct Option is D
Solution and Explanation
Answer (d) $x^2 + y^2 = 12$
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Top Questions on Hyperbola
- If the line \( ax + 2y = 1 \), where \( a \in \mathbb{R} \), does not meet the hyperbola \( x^2 - 9y^2 = 9 \), then a possible value of \( a \) is:
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- Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is:
- Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $ 2a $ and $ 2b $, respectively, and one focus and the corresponding directrix of this hyperbola be $ (-5, 0) $ and $ 5x + 9 = 0 $, respectively. If the product of the focal distances of a point $ (\alpha, 2\sqrt{5}) $ on the hyperbola is $ p $, then $ 4p $ is equal to:
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Questions Asked in KCET exam
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \).
Choose the correct answer from the options given below:- KCET - 2025
- Differentiability
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- Differentiability
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\[ f(x) = \begin{cases} \frac{1}{e^x - 1}, & \text{if } x \neq 0 \\ \frac{1}{e^x + 1}, & \text{if } x = 0 \end{cases} \] Then, which of the following is true?- KCET - 2025
- Limits
- The function f(x) is given by:
For x < 0:
f(x) = ex + axFor x ≥ 0:
f(x) = b(x - 1)2
The function is differentiable at x = 0. Then,- KCET - 2025
- Differentiability
- The function \( f(x) = \tan x - x \)
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- Derivatives
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
