TS EAMCET · Maths · Parabola
The equation of the given curve is \(x^2-4 x+4 y-8=0\) Match the following: List-I (A) Focus (B) Vertex (C) One end of the latus rectum (D) Point of intersection of the axis and directrix List-II (I) \((4,2)\) (II) \((3,2)\) (III) \((2,3)\) (IV) \((2,4)\) (V) \((2,2)\) The correct answer is \(\begin{array}{lllll} & \text { A } & \text { B } & \text { C } & \text { D } \end{array}\)
- A II III I IV
- B IV III I V
- C V III IV I
- D V III I IV
Answer & Solution
Correct Answer
(D) V III I IV
Step-by-step Solution
Detailed explanation
\(\mathrm{x}^2-4 \mathrm{x}+4 \mathrm{y}-8=0\) \[ \begin{aligned} & \Rightarrow \mathrm{x}^2-4 \mathrm{x}+4=-4 \mathrm{y}+8+4 \\ & \Rightarrow(\mathrm{x}-2)^2=-4(\mathrm{y}-3) \end{aligned} \] \(\therefore\) (B) vertex \(=(2,3)\) and axis \(=y\) - axis…
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