00:01
Hi, today we are solving the question in which we need to find all point on the curve given where the tangent line is horizontal for a.
00:09
So, xy is equal to 16y square plus x.
00:13
So, here to find where the tangent line is horizontal, we can take dy by dx and set it equal to zero.
00:24
So, here let us differentiate the given equation implicitly with respect to x.
00:30
So, d by dx of xy and d by dx of 16y square plus x.
00:37
Now, we get using the product rule y plus x into dy by dx is equal to 32y into y dash plus 1.
00:50
Now, let us solve for dy by dx when it is equal to zero.
00:55
So, dy by dx is equal to zero for tangent horizontal.
01:00
So, x into dy by dx is equal to 32y into y dash plus 1 minus y.
01:08
Now, from here, dy by dx, dy by dx can be taken as x minus 32y is equal to 1 minus y.
01:28
So, from here we can take dy by dx as dy by dx as equals to 1 minus y by x minus 32 into y setting it equal to zero.
01:42
We can write it as y is 1 minus y is equal to zero.
01:47
So, y is equals to minus 1, y is equals to 1.
01:52
Now, substituting this back into the original equation to find the corresponding x coordinate.
01:58
So, on substituting back, we can write it as x into 1 is equal to 16 into 1 square plus x.
02:11
So, x is equals to 16 plus x.
02:19
So, here we get the value of x as equals to x is equals to 16.
02:39
So, from here, therefore, required coordinates x or y will be equal to 16 or 1...