question_answer

Find the value of p for which the curves\[{{x}^{2}}-9p(9-y)\] and \[{{x}^{2}}=p(y\,+1)\] cut each other at right angle.

Answer:

Given equations of curves are \[{{x}^{2}}=9p(9-y)\]                         ?(i) and       \[{{x}^{2}}=p(y\,+1)\]                                   ?(ii) As these curves cut each other at right angles, therefore their tangent at point of intersection are perpendicular to each other. So, let us first find the point of intersection and slope of tangents to the curves. From Eqs. (i) and (ii), we get             \[9p(9-y)=p(y+1)\] \[\Rightarrow \]   \[9(9-y)=y+1\] [\[\because \,\,p\ne 0\,\,\text{as}\,\,\text{if}\,\,p=0,\] then curves become straight, which will be parallel] \[\Rightarrow \]   \[81-9y=y+1\] \[\Rightarrow \]   \[80=10y\] \[\Rightarrow \]   y = 8 On substituting the value of y in Eq. (i), we get             \[{{x}^{2}}=9p\] \[\Rightarrow \]   \[x=\pm \,3\sqrt{p}\] Thus, the point of intersection are \[(3\sqrt{p},\,\,8)\] and \[(-\,3\sqrt{p},\,\,8).\] Now, consider Eq. (i), we get             \[\frac{{{x}^{2}}}{9p}=9-y\] \[\Rightarrow \]   \[y=9-\frac{{{x}^{2}}}{9p}\] On differentiating bothe sides w.r.t. x, we get             \[\frac{dy}{dx}=\frac{-\,2x}{9p}\]                                  ?(iii) From Eq. (ii), we get                \[\frac{{{x}^{2}}}{p}=y+1\] \[\Rightarrow \] \[y=\frac{{{x}^{2}}}{p}-1\]                             On differentiating both sides w.r.t. x, we get           Now, at intersection point \[(3\sqrt{p},\,\,8),\] we have slope of tangent to first curve \[=\frac{-\,6\sqrt{p}}{9p}\]                                                 [using Eq. (iii)] and slope of tangent to second curve \[=\frac{6\sqrt{p}}{p}\] [using Eq. (iv)] \[\because \] Tangents are perpendicular to each other. \[\therefore \]      \[\frac{6\sqrt{p}}{9p}\cdot \frac{-\,6\sqrt{p}}{p}=-\,1\]      \[[\because \,\,\,{{m}_{1}}\cdot {{m}_{2}}=-\,1]\]\[\Rightarrow \]         \[\frac{4}{p}=1\] \[\Rightarrow \] p = 4                       Hence, the value of p is 4.