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Question : 152 of 160
Marks:
+1,
-0
Solution:
‌‌ Given, ‌z=tan(y+ax)+√y−ax‌⇒‌‌zx= sec2(y+ax)a+‌(−a)‌⇒‌‌zxx=2 sec2(y+ax)‌tan(y+ax)‌a2‌+‌‌‌ and ‌zy= sec2(y+ax)+‌‌⇒‌‌zyy=2 sec2(y+ax)‌tan(y+ax)‌−‌‌∴‌‌zxx−a2zyy=0‌
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