A) -64
B) 128
C) -32
D) -128
Correct Answer: D
Solution :
| [d] |
| Tangent on \[{{y}^{2}}=4x\text{ }is\text{ }y=mx+\frac{1}{m}\] |
| But given \[y=mx+4\] |
| \[\therefore \frac{1}{m}=4\] i.e. \[m=\frac{1}{4}\] |
| Now \[y=\frac{x}{4}+4\] is tangent on \[{{x}^{2}}=2by\] also |
| \[\therefore {{x}^{2}}=2b\left( \frac{x}{4}+4 \right)\] |
| \[{{x}^{2}}-\frac{bx}{2}-8b=0\] |
| For tangent Discriminant = 0 |
| \[\frac{{{b}^{2}}}{4}+32b=0\] |
| \[\Rightarrow b=-128\] |
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