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Find the point on the curve y^2=a x the ...

Find the point on the curve `y^2=a x` the tangent at which makes an angle of `45^0` with the x-axis.

Text Solution

Verified by Experts

The correct Answer is:
(a/4, a/2)
The equation of any tangent to the parabola `Y^(2)=4ax` in terms of its slope m is
`y=mx+(a)/(m)`
The coordinates of the point of contact are `(a//m^(2),2a//m)`.
Therefore, the equation of tangent to `y^(2)=ax` is
`y=mx+(a)/(4m)`
and the coordinates of the point of contact are `(a//4m^(2),a//2m)`.
It is given that `m=tan45^(@)=1`.
Therefore, the coordinates of the point of contact are `(a//4,a//2)`.
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Knowledge Check

  • The point on the curve y^(2)=x , the tangent at which makes an angle 45^(@) with the x- axis is-

    A
    `(0, 0)`
    B
    `((1)/(4), (1)/(2))`
    C
    `((1)/(2), (1)/(4))`
    D
    `(2, 4)`

  • The equation of the normal to the parabola y^(2)=5x . Which makes an angle of 45^(@) with the x-axis is -

    A
    `x-y=15`
    B
    `2(x-y)=15`
    C
    `4(x-y)=15`
    D
    `8(x-y)=15`

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