COMEDK · Maths · 13. Parabola
The ends of the latusrectum of the parabola \(x^{2}+10 x-16 y+25=0\) are
- A \((3,4),(-13,4)\)
- B \((5,-8),(-5,8)\)
- C \((3,-4),(13,4)\)
- D \((-3,4),(13,-4)\)
Answer & Solution
Correct Answer
(A) \((3,4),(-13,4)\)
Step-by-step Solution
Detailed explanation
Given, equation of parabola is \[ x^{2}+10 x-16 y+25=0 \] It can be rewritten as \[ (x+5)^{2}=16 y \] Here \(a=4\) and vertex \(\equiv(-5,0)\) and focus \(=(-5,4)\) So, \(y\)-coordinate of ends of latusrectum will be 4 . Putting in equation, we have \((x+5)^{2}=16 \times 4=64\)…
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