
The parabolas $y={{x}^{2}}-9$ and $y=k{{x}^{2}}$ intersect at a point A and B. If length AB is equal to 2a, then the value of k is $\dfrac{{{a}^{2}}-p}{{{a}^{2}}}$ , then find the value of p.
Answer
585k+ views
Hint: 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,
${{x}^{2}}-9=k{{x}^{2}}$
Solving for x we get,
${{x}^{2}}\left( 1-k \right)=9$
Taking the square root on both sides we get,
$x=\pm \sqrt{\dfrac{9}{1-k}}$
Substituting the value of ${{x}^{2}}$ in any of the parabolas.
We will substitute in the second parabola, we get,
$\begin{align}
& y=\dfrac{k\times 9}{1-k} \\
& =\dfrac{9k}{1-k} \\
\end{align}$
Now, we have got both the coordinates of A and B.
Therefore point A is $A\left( \sqrt{\dfrac{9}{1-k}},\dfrac{9k}{1-k} \right)$ and point be $B\left( -\sqrt{\dfrac{9}{1-k}},\dfrac{9k}{1-k} \right)$ .
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 = $\left| {{x}_{2}}-{{x}_{1}} \right|$ ,
Substituting the values, we get,
$2a=2\sqrt{\dfrac{9}{1-k}}$
Solving for k, we get,
$a=\sqrt{\dfrac{9}{1-k}}$
Squaring on both sides we get,
${{a}^{2}}=\dfrac{9}{1-k}$
$1-k=\dfrac{9}{{{a}^{2}}}$
$\begin{align}
& k=1-\dfrac{9}{{{a}^{2}}} \\
& =\dfrac{{{a}^{2}}-9}{{{a}^{2}}}
\end{align}$
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.
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,
${{x}^{2}}-9=k{{x}^{2}}$
Solving for x we get,
${{x}^{2}}\left( 1-k \right)=9$
Taking the square root on both sides we get,
$x=\pm \sqrt{\dfrac{9}{1-k}}$
Substituting the value of ${{x}^{2}}$ in any of the parabolas.
We will substitute in the second parabola, we get,
$\begin{align}
& y=\dfrac{k\times 9}{1-k} \\
& =\dfrac{9k}{1-k} \\
\end{align}$
Now, we have got both the coordinates of A and B.
Therefore point A is $A\left( \sqrt{\dfrac{9}{1-k}},\dfrac{9k}{1-k} \right)$ and point be $B\left( -\sqrt{\dfrac{9}{1-k}},\dfrac{9k}{1-k} \right)$ .
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 = $\left| {{x}_{2}}-{{x}_{1}} \right|$ ,
Substituting the values, we get,
$2a=2\sqrt{\dfrac{9}{1-k}}$
Solving for k, we get,
$a=\sqrt{\dfrac{9}{1-k}}$
Squaring on both sides we get,
${{a}^{2}}=\dfrac{9}{1-k}$
$1-k=\dfrac{9}{{{a}^{2}}}$
$\begin{align}
& k=1-\dfrac{9}{{{a}^{2}}} \\
& =\dfrac{{{a}^{2}}-9}{{{a}^{2}}}
\end{align}$
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

