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Question: Equation of a common tangent to the parabola \({{y}^{2}}=4x\) and the hyperbola xy = 2 is: A. x + ...

Equation of a common tangent to the parabola y2=4x{{y}^{2}}=4x and the hyperbola xy = 2 is:
A. x + 2y + 4 = 0
B. x – 2y + 4=0
C. x + y +1 = 0
D. 4x + 2y + 1 = 0

Explanation

Solution

Assume an equation for tangent to the parabola y2=4x{{y}^{2}}=4x with some unknown variable ‘m’ such as y=mx+1my=mx+\dfrac{1}{m}. Substitute the equation in the given equation of hyperbola xy = 2. Solve it using the equation to find the root of a quadratic equation and find the value of ‘m’. Here, we have to take the determinant to be zero. Then substitute it in the assumed equation.

Complete step by step answer:
First of all we have to assume an equation for the tangent to the parabola y2=4x{{y}^{2}}=4x.
Let the equation be y=mx+1my=mx+\dfrac{1}{m}.
Now, we have to substitute the value of y in the given equation of hyperbola. So we get,
xy = 2
\Rightarrow x×(mx+1m)=2x\times \left( mx+\dfrac{1}{m} \right)=2
On solving, we get,
mx2+xm=2m{{x}^{2}}+\dfrac{x}{m}=2
Now, we have to rearrange the equation,
mx2+xm2=0\Rightarrow m{{x}^{2}}+\dfrac{x}{m}-2=0
m2x2+x2m=0\Rightarrow {{m}^{2}}{{x}^{2}}+x-2m=0
Solve the above quadratic equation, by taking the value of the determinant as,
D = 0.
Therefore,b24ac=0{{b}^{2}}-4ac=0
So the equation becomes,
12(4×m2×(2m))=0{{1}^{2}}-(4\times {{m}^{2}}\times (-2m))=0
1+8m3=0\Rightarrow 1+8{{m}^{3}}=0
On further solving, we get
8m3=18{{m}^{3}}=-1
m3=18\Rightarrow {{m}^{3}}=\dfrac{-1}{8}
Now, we can take the cube root of ‘m’, so we get,
m=12m=\dfrac{-1}{2}
As we got the value of ‘m’, now we have to substitute it in the assumed equation for the tangent of parabola y2=4x{{y}^{2}}=4x
So, the equation y=mx+1my=mx+\dfrac{1}{m} becomes,
y=12x+1(12)y=\dfrac{-1}{2}x+\dfrac{1}{\left( \dfrac{-1}{2} \right)}
y=x22\Rightarrow y=\dfrac{-x}{2}-2
2y=x4\Rightarrow 2y=-x-4
Now, we have to rearrange the equation to get the final answer.
x+2y+4=0x+2y+4=0

So, the correct answer is “Option A”.

Note: We should be careful while assuming an equation for the tangent of the parabola y2=4x{{y}^{2}}=4x. The discriminant of the quadratic equation must be 0. The discriminant is zero for one real solution, that is there will be one real root. So, use the equation D=b24acD={{b}^{2}}-4ac. Solve the equation to the simplest form.