AP EAMCET · Maths · Functions
If \(a \neq 0\) and the line \(2 b x+3 c y+4 d=0\), passes through the points of intersection of the parabolas \(y^2=4 a x\) and \(x^2=4 a y\), then
- A \(d^2+(2 b+3 c)^2=0\)
- B \(d^2+(3 b+2 c)^2=0\)
- C \(d^2+(2 b-3 c)^2=0\)
- D \(d^2+(3 b-2 c)^2=0\)
Answer & Solution
Correct Answer
(A) \(d^2+(2 b+3 c)^2=0\)
Step-by-step Solution
Detailed explanation
\(y^2=4 a x, x^2=4 a y\) clearly they are symmetric w.r.t \(y=x\). \(\therefore y=x\) also pass through their point of intersections. i.e., solving \(y^2=4 a x\) and \(y=x\) \(\Rightarrow x^2=4 a x \Rightarrow x=0,4 a\) When \(x=0, y=0 ; x=4 a \Rightarrow y=4 a\)…
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