Question

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectus of each parabola is 3, then the equation of the common tangent to the two parabolas is:
A. \[3\left( x+y \right)+4=0\]
B. \[8\left( 2x+y \right)+3=0\]
C. \[4\left( x+y \right)+3=0\]
D. \[x+2y+3=0\]

Answer 1

Hint: First we need to find out the common vertex i.e. Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. Therefore, the only point common to x-axis and the y-axis is the origin i.e. \[\left( 0,0 \right)\]. Then we will let the equation of the two parabolas i.e. \[{{y}^{2}}=4ax\] and\[{{x}^{2}}=4by\]. Now putting the value of latus rectum i.e. equal to 3 in the equations, we will get \[{{y}^{2}}=3x\] and \[{{x}^{2}}=3y\]. Then we let the common tangent be \[y=mx+c\]. And substituting this value in each of the parabola equations and solving them we will get the value of ‘m’ and ‘c’. Substituting it in the equation of the tangent, we will get our required answer.

Formula used:
Equation of parabola is of the form;
\[ {{y}^{2}}=4ax\] and \[{{x}^{2}}=4by\].
Equation of the tangent is of the form \[y=mx+c\]

Complete step by step solution:
We have given that,
Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant.
Therefore,
The only point common to x-axis and the y-axis is the origin i.e. \[\left( 0,0 \right)\].
Thus,
\[\Rightarrow Origin\ \left( 0,0 \right)\] is the common vertex.
Let the equation of the two parabolas be;
\[\Rightarrow {{y}^{2}}=4ax\]
And
\[\Rightarrow {{x}^{2}}=4by\]
It is given that the length of the latus rectus of each parabola is 3.
Latus rectum = 3
Thus,
\[\Rightarrow 4a=4b=3\]
So,
\[\Rightarrow a=b=\dfrac{3}{4}\]
Now,
The equation of the two parabolas are;
\[\Rightarrow {{y}^{2}}=3x\]
And
\[\Rightarrow {{x}^{2}}=3y\]
Now,
Let the common tangent be \[y=mx+c\]
We have,
\[\Rightarrow {{y}^{2}}=3x\]
Substituting the value of \[y=mx+c\]
\[\Rightarrow {{\left( mx+c \right)}^{2}}=3x\]
Using the identity of\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
\[\Rightarrow {{m}^{2}}{{x}^{2}}+2mxc+{{c}^{2}}=3x\]
Subtracting 3x from both the sides, we will get
\[\Rightarrow {{m}^{2}}{{x}^{2}}+2mxc+{{c}^{2}}-3x=0\]
Taking the common terms, we will get
\[\Rightarrow {{m}^{2}}{{x}^{2}}+x\left( 2mc-3 \right)+{{c}^{2}}=0\]
As we know that,
The tangent touches at only one point where the value of discriminant is equal to zero.
Thus,
\[{{b}^{2}}-4ac=0\]
Here, \[a={{m}^{2}},\ b=2mc-3,\ c={{c}^{2}}\]
Substituting the values, we will get
\[\Rightarrow {{\left( 2mc-3 \right)}^{2}}-4{{m}^{2}}{{c}^{2}}=0\]
Expanding the above using the identity of\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].
\[\Rightarrow 4{{m}^{2}}{{c}^{2}}-12mc+9-4{{m}^{2}}{{c}^{2}}=0\]
Combining the like terms, we will get
\[\Rightarrow -12mc+9=0\]
Solving for the value of ‘c’, we will get
\[\Rightarrow c=\dfrac{3}{4m}\]------ (1)
And
We have,
\[\Rightarrow {{x}^{2}}=3y\]
Substituting the value of \[y=mx+c\]
\[\Rightarrow {{x}^{2}}=3\left( mx+c \right)\]
Simplifying the above we will get,
\[\Rightarrow {{x}^{2}}-3mx-3c=0\]
As we know that,
The tangent touches at only one point where the value of discriminant is equal to zero.
Thus,
\[{{b}^{2}}-4ac=0\]
Here, \[a=1,\ b=-3m,\ c=-3c\]
Substituting the values, we will get
\[\Rightarrow {{\left( -3m \right)}^{2}}-4\left( 1 \right)\left( -3c \right)=0\]
Simplifying the above, we will get
\[\Rightarrow 9{{m}^{2}}+12c=0\]
Substituting the value of ‘c’ from the equation (1),
\[\Rightarrow 9{{m}^{2}}+12\times \dfrac{3}{4m}=0\]
\[\Rightarrow 9{{m}^{2}}+\dfrac{9}{m}=0\]
Taking the LCM of the denominators, we will get
\[\Rightarrow \dfrac{9{{m}^{3}}}{m}+\dfrac{9}{m}=0\]
Multiplying both the sides of the equation by ‘m’,
\[\Rightarrow 9{{m}^{3}}+9=0\]
Subtracting 9 form both the sides of the equation, we will get
\[\Rightarrow 9{{m}^{3}}=-9\]
Dividing both the sides of the equation by 9, we will get
\[\Rightarrow {{m}^{3}}=-1\]
Thus,
\[\Rightarrow m=-1\]
Now,
We have,
\[\Rightarrow c=\dfrac{3}{4m}=-\dfrac{3}{4}\]
Therefore,
The equation of tangent is given by;
\[\Rightarrow y=mx+c\]
Here, \[m=-1\ and\ c=-\dfrac{3}{4}\]
Substituting the values, we will get
\[\Rightarrow y=-x-\dfrac{3}{4}\]
Or
\[\Rightarrow y=-x-\dfrac{3}{4}\Rightarrow 4y=-4x-3\Rightarrow 4x+4y+3\Rightarrow 4\left( x+y \right)+3\]
\[\Rightarrow 4\left( x+y \right)+3=0\]
Hence, the equation of the tangent is \[4\left( x+y \right)+3=0\].

Hence, the option (C) is the correct answer.

Note: Whenever we face such a type of question we have to assume the tangent equation in the slope form and equate it with the given equation to find out the value of slope that we assumed in the tangent equation. And then putting the value of slope and the constant we will get the required tangent equation. Tangent equation in the slope form is given by\[y=mx+c\], where ‘m’ is the slope of the equation.