JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(P\) be a variable point on the parabola \(y=4 x^{2}+1\). Then the locus of the mid-point of the point \(P\) and the foot of the perpendicular drawn from the point \(P\) to the line \(y=x\) is:
- A \((3 x-y)^{2}+(x-3 y)+2=0\)
- B \(2(x-3 y)^{2}+(3 x-y)+2=0\)
- C \(2(3 x-y)^{2}+(x-3 y)+2=0\)
- D \((3 x-y)^{2}+2(x-3 y)+2=0\)
Answer & Solution
Correct Answer
(C) \(2(3 x-y)^{2}+(x-3 y)+2=0\)
Step-by-step Solution
Detailed explanation
\(\frac{K-C}{h-C}=-1\) \(C=\frac{h+k}{2} \quad P(x, y)\) \(R=\left(\frac{x+C}{2}, \frac{y+C}{2}\right)\) \(R=\left(\frac{x}{2}+\frac{h}{4}+\frac{K}{4}, \frac{y}{2}+\frac{h}{4}+\frac{k}{4}\right)\) \(\therefore x=\frac{3 h}{2}-\frac{K}{2}, y=\frac{3 K}{2}-\frac{h}{2}\), put in…
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