Find the equation of the tangent to the parabola `x=y^2+3y+2` having slope 1.
Find the equation of the tangent to the parabola `x=y^2+3y+2` having slope 1.
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`x=y^(2)+3y+2`
Differentiating both sides w.r.t x, we have
`1=2y(dy)/(dx)+3(dy)/(dx)`
`or(dy)/(dx)=(1)/(2y+3)`
Now, the slope of the tangent is 1. Therefore,
`or(dy)/(dx)=(1)/(2y+3)=1`
`ory=-1`
which is the ordinate of the point on the curve where the slope of the tangent is 1.
Putting y=-1 in the equation of parabola, we get x=0.
Hence, using-form, we have
`y-(-1)=1(x-0)`
`orx-y-1=0`
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