Question
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 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
Solution
Assume an equation for tangent to the parabola y2=4x with some unknown variable ‘m’ such as y=mx+m1. 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.
Let the equation be y=mx+m1.
Now, we have to substitute the value of y in the given equation of hyperbola. So we get,
xy = 2
⇒ x×(mx+m1)=2
On solving, we get,
mx2+mx=2
Now, we have to rearrange the equation,
⇒mx2+mx−2=0
⇒m2x2+x−2m=0
Solve the above quadratic equation, by taking the value of the determinant as,
D = 0.
Therefore,b2−4ac=0
So the equation becomes,
12−(4×m2×(−2m))=0
⇒1+8m3=0
On further solving, we get
8m3=−1
⇒m3=8−1
Now, we can take the cube root of ‘m’, so we get,
m=2−1
As we got the value of ‘m’, now we have to substitute it in the assumed equation for the tangent of parabola y2=4x
So, the equation y=mx+m1 becomes,
y=2−1x+(2−1)1
⇒y=2−x−2
⇒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 y2=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=b2−4ac. Solve the equation to the simplest form.