Question
The equation of the tangent to the parabola y2 = 4x at the point (1, 2) is-
- x – y + 1 = 0
- x + y – 1 = 0
- x + y + 1 = 0
- x – y – 1 = 0
Hint:
find the slope of the curve at the given point and find the equation of straight line
The correct answer is: x – y + 1 = 0
x – y + 1 = 0
Slope of tangent = dy/dx = 2/y
Slope at 1,2 = 1
Equation of tangent : y-2 = 1(x-1)
y = x+1
x – y + 1 = 0
.
Slope of tangent = dy/dx
Equation of line: y-y1 = m(x-x1)
Related Questions to study
If a tangent to the parabola 4y2 = x makes an angle of 60º with the x- axis, then its point of contact is-
point of contact = (a/m2, 2a/m)
slope = tan ( angle made with the x axis)
this gives us the value of m
y^2 = x/4 = 4(1/16)x
a = 1/16
If a tangent to the parabola 4y2 = x makes an angle of 60º with the x- axis, then its point of contact is-
point of contact = (a/m2, 2a/m)
slope = tan ( angle made with the x axis)
this gives us the value of m
y^2 = x/4 = 4(1/16)x
a = 1/16
At which point the line x = my + touches the parabola x2 =4ay
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At which point the line x = my + touches the parabola x2 =4ay
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For what value of k, the line 2y – x + k = 0 touches the parabola x2 + 4y = 0-
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For what value of k, the line 2y – x + k = 0 touches the parabola x2 + 4y = 0-
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a quadratic equation has equal roots if the D value is zero. this is the condition of the line touching the curve.
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the solution of the two curves gives us the quadratic equation. D=0 gives us the required answer.
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the solution of the two curves gives us the quadratic equation. D=0 gives us the required answer.
The straight line x + y = k touches the parabola y = x – x2, if k =
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The straight line x + y = k touches the parabola y = x – x2, if k =
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