Equation of common tangent of y=x^2,y=-...

Equation of common tangent of `y=x^2,y=-x^2+4x-4` is

y = 4 (x - 1)

y = 0

y = - 4(x - 1)

y = - 30x - 50

Text Solution

Verified by Experts

The correct Answer is:

A, B

The equation of tangent to `y = x^(2)`, be `y = mx - (m^(2))/(4)`
Putting in `y = - x^(2) +4 x - 4` we should only get one value of x i.e., Discrimunant must be zero.
`:.mx - (m^(2))/(4) = - x^(2) + 4x - 4`
`implies x^(2) + x (m - 4) + 4 - (m^(4))/(4) = 0`
Now, `(m - 4)^(2) - (16 - m^(2)) = 0`
`implies 2m (m - 4) = 0 implies m = 0,4`
`:.` y = 0 and y = 4 (x - 1) are the requried tangents.
Hence, (a) and (b) are correct answers.

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The common tangent(s) of y=x^2 and y=-x^2 + 4x – 4 is (are) :

`y=4(x - 1)`

`y=- 4(x - 1)`

`y=0 `

`y=30x-20`

The equation of common tangent to the curve y^2=4x and xy=-1 is

`2^(2/3)*y=x+2^(4/3)`

`2^(2/3)*x=y+2^(4/3)`

`y=2^(2/3)*x-2^(4/3)`

`y=x-2^(2/3)`

Equation of a common tangent to the parabola y^(2)=4x and the hyperbola xy=2 is

x+2y+4=0

x-2y+4=0

x+y+1=0

4x+2y+1=0