JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The distance of the point \((6,-2 \sqrt{2})\) from the common tangent \(y = mx + c , m > 0\), of the curves \(x =2 y ^2\) and \(x =1+ y ^2\) is
- A \(\frac{1}{3}\)
- B \(5\)
- C \(\frac{14}{3}\)
- D \(5 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
For \(y ^2=\frac{ x }{2}, T : y = mx +\frac{1}{8 m }\) For tangent to \(y ^2+1= x\) \(\Rightarrow\left( mx +\frac{1}{8 m }\right)^2+1= x\) \(D =0 \Rightarrow m =\frac{1}{2 \sqrt{2}}\) \(\therefore T : x -2 \sqrt{2} y +1=0\) \(d =\left|\frac{6+8+1}{\sqrt{9}}\right|=5\)
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