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Two tangents are drawn from the point (-2,-1) to the parabola y2=4x. If α is the angle between them, then tan α =
- a)3
- b)1/3
- c)2
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Two tangents are drawn from the point (-2,-1) to the parabola y2=4x. I...
The eqn of any tangent to a parabola y^2 = 4ax is y = mx+a/m
So the equation of any tangent to this parabola, would be y=mx+1/m (as a=1, for y^2=4x)
Now these tangents pass through (-2,-1)... so -1=-2m+1/m
Or 2m2-m-1=0.
Where m1,m2 are the slopes of the two tangents that satisfy this equation (m1,m2 are the roots of the quad)....
So tan(angle between the tangents) = |(m1-m2)/(1+m1m2)|
Now m1m2 = -1/2
And m1-m2 = root { (m1+m2)2-4m1m2 } = root [ (1/2)2 - 4(-1/2) ] = root[9/4].
So |m1-m2| = 3/2.
Hence tan(Angle) = (3/2) / (1/2) = 3
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Two tangents are drawn from the point (-2,-1) to the parabola y2=4x. I...
We draw a diagram, we can see that the two tangents will intersect at a point on the parabola.
Let's find the equation of the tangent line to the parabola at the point (a,b).
Using the equation of the parabola, we can find the slope of the tangent line at any point (a,b) by taking the derivative of y^2=4x with respect to x.
Differentiating both sides of the equation, we get:
2y(dy/dx) = 4
dy/dx = 4/(2y)
At the point (a,b), we substitute b for y and get:
dy/dx = 4/(2b) = 2/b
Since the slope of the tangent line is the negative reciprocal of the slope of the radius (which is 1/2), the slope of the tangent line is -2.
Using the point-slope form of a line, we can write the equation of the tangent line as:
y - b = -2(x - a)
We know that the point (-2,-1) lies on the tangent line, so we substitute these values into the equation:
-1 - b = -2(-2 - a)
-1 - b = 4 + 2a
b = -2a - 5
Substituting this expression for b into the equation of the tangent line, we get:
y - (-2a - 5) = -2(x - a)
y + 2a + 5 = -2x + 2a
y = -2x
So, the equation of the tangent line is y = -2x.
To find the points of intersection between the tangent lines and the parabola, we substitute y = -2x into the equation of the parabola:
(-2x)^2 = 4x
4x^2 = 4x
4x^2 - 4x = 0
4x(x - 1) = 0
x = 0 or x = 1
If x = 0, then y = -2(0) = 0, so one point of intersection is (0,0).
If x = 1, then y = -2(1) = -2, so the other point of intersection is (1,-2).
Therefore, the two tangents from the point (-2,-1) to the parabola y^2 = 4x intersect the parabola at the points (0,0) and (1,-2).
Let's find the equation of the tangent line to the parabola at the point (a,b).
Using the equation of the parabola, we can find the slope of the tangent line at any point (a,b) by taking the derivative of y^2=4x with respect to x.
Differentiating both sides of the equation, we get:
2y(dy/dx) = 4
dy/dx = 4/(2y)
At the point (a,b), we substitute b for y and get:
dy/dx = 4/(2b) = 2/b
Since the slope of the tangent line is the negative reciprocal of the slope of the radius (which is 1/2), the slope of the tangent line is -2.
Using the point-slope form of a line, we can write the equation of the tangent line as:
y - b = -2(x - a)
We know that the point (-2,-1) lies on the tangent line, so we substitute these values into the equation:
-1 - b = -2(-2 - a)
-1 - b = 4 + 2a
b = -2a - 5
Substituting this expression for b into the equation of the tangent line, we get:
y - (-2a - 5) = -2(x - a)
y + 2a + 5 = -2x + 2a
y = -2x
So, the equation of the tangent line is y = -2x.
To find the points of intersection between the tangent lines and the parabola, we substitute y = -2x into the equation of the parabola:
(-2x)^2 = 4x
4x^2 = 4x
4x^2 - 4x = 0
4x(x - 1) = 0
x = 0 or x = 1
If x = 0, then y = -2(0) = 0, so one point of intersection is (0,0).
If x = 1, then y = -2(1) = -2, so the other point of intersection is (1,-2).
Therefore, the two tangents from the point (-2,-1) to the parabola y^2 = 4x intersect the parabola at the points (0,0) and (1,-2).
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Two tangents are drawn from the point (-2,-1) to the parabola y2=4x. If α is the angle between them, then tan α =a)3b)1/3c)2d)1/2Correct answer is option 'A'. Can you explain this answer?
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