Given that tangent P (y=px-ap^2) and tanget Q (y=qx-aq^2) intersect at an angle 60 degress. Show that p-q=sqaureroot 3 (1+pq)Two points P(2ap, ap^2) and Q(2aq, aq^2) lie on the parabola x^2=4ay

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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Just look at the example lines shown in the figure.

Let us say the angle between x axis and the line black line is `alpha` .

Let us say the angle between x axis and the line red line is `beta` .

Let us say the small angle of intersection is `theta` .

 

Now if you apply the exterior angle theorem between `alpha` ,`beta` and `theta ` you will get;

`theta+alpha = beta`

Or in other words;

`theta = alpha-beta`

 

Let us say m1 is the gradient of the black line.

Let us say m2 is the gradient of the red line.

 

Then;

tanalpha = m1 --->`alpha = tan^(-1)(m1)`

tanbeta = m2 ---->`beta = tan^(-1)(m2)`

 

Therefore;

`theta = tan^(-1)(m1)-tan^(-1)(m2) ` ----(1)

 

Now move on to our question.

The gradients of the P and Q lines are p and q respectively.

So;

`m1 = p`

`m2 = q`

`theta = 60`

 

so applying this data to (1) yields;

`60 = tan^(-1)(p)-tan^(-1)(q)`

Get the tan in both sides.

 

`tan60 = tan(tan^(-1)(p)-tan^(-1)(q))`

 

We know that;

`tan(A-B) = (tanA-tanB)/(1+tanAtanB)`

 

So;

`tan60 = [tan(tan^(-1)(p))-tan(tan^(-1)(q))]/(1+tan(tan^(-1)(p))*tan(tan^(-1)(q)))`

`sqrt3 = (p-q)/(1+pq)`

 

`sqrt3*(1+pq) = p-q`

Sources:

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