| Direction: A straight line will touch a given conic if there is only one point of intersection of the line and the given conic. If the conic is specified by quadratic equation in\[x\] and \[y,\] then the straight line will touch if the discriminant of the equation obtained by the elimination of one of the variable is zero. Let us consider parabola \[{{y}^{2}}=8x\] and an ellipse\[15{{x}^{2}}+4{{y}^{2}}=60\]. |
A) \[2x-y-8=0\]
B) \[x-2y+8=0\]
C) \[2x-y+8=0\]
D) \[x+2y-8=0\]
Correct Answer: B
Solution :
Let the tangent of \[{{y}^{2}}=8x\] is \[y=mx+2\,/m\] and the tangent of \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{15}=1\] is \[y=mx\,\pm \,\sqrt{4{{m}^{2}}+15}\] From Eqs. (i) and (ii), we get \[\frac{2}{m}=\pm \,\sqrt{4{{m}^{2}}+15}\] \[\Rightarrow \] \[4{{m}^{4}}+15{{m}^{2}}-4=0\] \[\Rightarrow \] \[m=\pm \,\frac{1}{2}\] Hence, equation of tangent is \[atm=\frac{1}{2}\] \[\Rightarrow \] \[y=\frac{x}{2}+4\] \[\Rightarrow \] \[2y=x+8\]
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