Jodis

Please: Differential equation (dy/dx)^2 + sin(x)cos^2(x)dy/dx - sin^4(x)=0 Answer is y=cos(x)+c or y=1/3 cos^3(x) - cos(x) + c but stuck. Thanks

The curve y=x^3-3x^2-8x+4 has tangent L at point P (1,-8). Given that the Line M is parallel to L and is also a tangent to Q show that the shortest distance between L and M is 16 root 2 Thanks

The curve y=x^3-3x^2-8x+4 has tangent L at point P (-1,8). Given that the Line M is parallel to L and is also a tangent to Q show that the shortest distance between L and M is 16 root 2 Sorry This is the correct question

Thanks I got cos(x)+c which matches my working. However, can you explain lines 3 and 4 i.e. sin^2x(1+sin^2x)^2/4 ---> ±sinx(1+sin^2x)/2

So for line 3... can I use (sin(x) (1+sin^2x)/2)^2 instead of sin^2x(1+sin^2x)^2/4 and then root the answer to get ±sin(x)(1+sin^2x)/2 Thanks again

I found the x for Q to be 3 and subbing x into the equation found y-coordinate as -20 Q(3,-20) P(-1,8) However using the distance equation I get 20root2 not 16root2. Can you check whether my working is right?

Thanks, I calculated both tangents but was stuck on the next part. Thanks for explanation