A) \[2x-y+1=0\]
B) \[2x+y+1=0\]
C) \[x-2y+8=0\]
D) Both &
Correct Answer: D
Solution :
| [d] We know the tangent to the parabola |
| \[{{y}^{2}}=4ax\] at \[(a{{t}^{2}},2at)\]is \[ty=x+a{{t}^{2}}.\] Here \[a=2\] so, the tangent at \[(2{{t}^{2}},4t)\] to the parabola |
| \[{{y}^{2}}=8x\] is \[ty=x+2{{t}^{2}}\] ?. (i) |
| ?m? of (i) is \[\frac{1}{t};\] (i) makes \[45{}^\circ \] with \[y=3x+5\]if |
| \[\tan 45{}^\circ =\left| \frac{\frac{1}{t}-3}{1+\frac{1}{t}.3} \right|=\left| \frac{1-3t}{t+3} \right|\] |
| \[\therefore 1=\left| \frac{1-3t}{t+3} \right|;\] Or \[\frac{1-3t}{t+3}=\pm 1;\] or |
| \[1-3t=t+3,-t-3\] |
| \[\therefore 4t=-2\] Or \[2t=4.\] \[\therefore t=-\frac{1}{2}or2\] |
| Putting in (i), the tangents have the equations |
| \[-\frac{1}{2}y=x+2.\frac{1}{4}i.e.,2x+y+1=0\] |
| and \[2y=x+2.i.e.,x-2y+8=0\] |
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