The equation of the curve

Question

The tangent at !!(x,y)!! intersects the !!y!! axis at !!(0,x^2)!!. And the curve passes through !!(1,1)!! and !!(2,t)!!. !!t!! is?

  • !!0!!
  • !!1/2!!
  • !!1!!
  • !!2!!

Solution

The equation of a tangent at !!(x_1,y_1)!! be !!y=mx+c!!

Here !!m=(dy)/(dx)!! and !!c=x_1^2!!

So !!y=(dy)/(dx)|_(x_1,y_1) x + x1^2!!

Now since it passes through !!(x_1,y_1)!! we get

!!y=(dy)/(dx) x+x^2!!

!!dy/dx-y/x= -x!!

!!I.F=e^(int -1/x dx)= 1/x!!

!!y*1/x=int-x*1/xdx!!

!!y*1/x=-x+c!!

!!y=-x^2+cx!!

!!Since it passes through !!(1,1)!! we get !!c=2!!

!!y=-x^2+2c!!

So the curve passes through !!2,0!!

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