Explanation
Hence, Option 'D' is correct
At any point (k,k2+8) on y=x2+8, the slope of tangent is dydx|x=k=2k.....(1)
Similarly, at any point (p,−p2) on y=−x2, the slope of tangent is dydx|x=p=−2p....(2)
For common tangent, 2k=−2p⇒k=−p
Hence, the coordinates of point on common tangent and on y=−x2 can be written as (−k,−k2) as shown in figure.
So, slope of common tangent is 8+k2−(−k2)k−(−k)=8+2k22k
From (1), slope is also equal to 2k.
Hence, 8+2k22k=2k
Solving, we get k=±2.
Discarding negative value (since 'k' is positive as defined in the function), we get k=2
Hence, slope of common tangent is 2k=4
The point common to tangent and y=x2+8 is (2,12)
Hence, equation of common tangent is y−12=4(x−2)
To find x intercept, substitute y=0 in the above equation.
We get x intercept=−1
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