Question

Equation of a common tangent to the parabola ${{y}^{2}}=4x$ 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

Answer 1

Hint: Assume an equation for tangent to the parabola ${{y}^{2}}=4x$ with some unknown variable ‘m’ such as \[y=mx+\dfrac{1}{m}\]. 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 ${{y}^{2}}=4x$.
Let the equation be \[y=mx+\dfrac{1}{m}\].
Now, we have to substitute the value of y in the given equation of hyperbola. So we get,
xy = 2
$\Rightarrow $$x\times \left( mx+\dfrac{1}{m} \right)=2$
On solving, we get,
$m{{x}^{2}}+\dfrac{x}{m}=2$
Now, we have to rearrange the equation,
$\Rightarrow m{{x}^{2}}+\dfrac{x}{m}-2=0$
$\Rightarrow {{m}^{2}}{{x}^{2}}+x-2m=0$
Solve the above quadratic equation, by taking the value of the determinant as,
D = 0.
Therefore,${{b}^{2}}-4ac=0$
 So the equation becomes,
${{1}^{2}}-(4\times {{m}^{2}}\times (-2m))=0$
$\Rightarrow 1+8{{m}^{3}}=0$
On further solving, we get
$8{{m}^{3}}=-1$
$\Rightarrow {{m}^{3}}=\dfrac{-1}{8}$
Now, we can take the cube root of ‘m’, so we get,
$m=\dfrac{-1}{2}$
As we got the value of ‘m’, now we have to substitute it in the assumed equation for the tangent of parabola ${{y}^{2}}=4x$
So, the equation $y=mx+\dfrac{1}{m}$ becomes,
$y=\dfrac{-1}{2}x+\dfrac{1}{\left( \dfrac{-1}{2} \right)}$
$\Rightarrow y=\dfrac{-x}{2}-2$
$\Rightarrow 2y=-x-4$
Now, we have to rearrange the equation to get the final answer.
$x+2y+4=0$

So, the correct answer is “Option A”.

Note: We should be careful while assuming an equation for the tangent of the parabola ${{y}^{2}}=4x$. 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 $D={{b}^{2}}-4ac$. Solve the equation to the simplest form.