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Question: The parabolas \(y={{x}^{2}}-9\) and \(y=k{{x}^{2}}\) intersect at a point A and B. If length AB is e...

The parabolas y=x2−9y={{x}^{2}}-9 and y=kx2y=k{{x}^{2}} intersect at a point A and B. If length AB is equal to 2a, then the value of k is a2−pa2\dfrac{{{a}^{2}}-p}{{{a}^{2}}} , then find the value of p.

Explanation

Solution

In order to solve this problem, we need to find the point of intersection of two parabolas and the finding the distance between the points with the help of difference in coordinates. If any coordinates of both the points are the same then the difference in the other two points will give the distance between these two points.

Complete step-by-step answer:
We need to start by finding the point of intersection of two parabolas
The points at which they intersect are A and B.
Both parabolas satisfy the point,
That means that we can write it as,
x2−9=kx2{{x}^{2}}-9=k{{x}^{2}}
Solving for x we get,
x2(1−k)=9{{x}^{2}}\left( 1-k \right)=9
Taking the square root on both sides we get,
x=±91−kx=\pm \sqrt{\dfrac{9}{1-k}}
Substituting the value of x2{{x}^{2}} in any of the parabolas.
We will substitute in the second parabola, we get,
y=k×91−k =9k1−k \begin{aligned} & y=\dfrac{k\times 9}{1-k} \\\ & =\dfrac{9k}{1-k} \\\ \end{aligned}
Now, we have got both the coordinates of A and B.
Therefore point A is A(91−k,9k1−k)A\left( \sqrt{\dfrac{9}{1-k}},\dfrac{9k}{1-k} \right) and point be B(−91−k,9k1−k)B\left( -\sqrt{\dfrac{9}{1-k}},\dfrac{9k}{1-k} \right) .
As we can see that the y coordinate is the same therefore the line joining AB is horizontal.
Therefore, the line is parallel to the x-axis.
We have been the distance between two points which is 2a.
We will only vary x coordinate because y coordinate is constant
Therefore distance = ∣x2−x1∣\left| {{x}_{2}}-{{x}_{1}} \right| ,
Substituting the values, we get,
2a=291−k2a=2\sqrt{\dfrac{9}{1-k}}
Solving for k, we get,
a=91−ka=\sqrt{\dfrac{9}{1-k}}
Squaring on both sides we get,
a2=91−k{{a}^{2}}=\dfrac{9}{1-k}
1−k=9a21-k=\dfrac{9}{{{a}^{2}}}
k=1−9a2 =a2−9a2\begin{aligned} & k=1-\dfrac{9}{{{a}^{2}}} \\\ & =\dfrac{{{a}^{2}}-9}{{{a}^{2}}} \end{aligned}
Therefore, with comparing the values from the question the value of p is 9.

Note: While taking the distance we took the modulus sign as the distance is always positive. Also, while taking square roots we need to take both the roots as that’s the only way we will be getting the two points of intersection. Also, as the y coordinate is constant therefore, the line passing through A and B is parallel to x-axis so the distance between them is the difference of x coordinates.