AP EAMCET · Maths · Parabola
The common tangent to the parabola \(y^2=32 x\) and \(x^2=256 y\) will be
- A \(2 x+4 y+64=0\)
- B \(x+2 y-32=0\)
- C \(2 x+4 y+32=0\)
- D \(4 x+2 y+64=0\)
Answer & Solution
Correct Answer
(A) \(2 x+4 y+64=0\)
Step-by-step Solution
Detailed explanation
Given parabolas are, \(y^2=32 x \text { and } x^2=256 y\) We use a standard result to find equation of common tangent. Equation of tangent common to \(y^2=4 a x\) and \(x^2=4 b y\) is, \(b^{1 / 3} y+a^{1 / 3} x+\left(a^2 b^2\right)^{1 / 3}=0\) Here,…
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