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Question

# The sum of x−intercept and y−intercept of the common tangent to the parabola y2=16x and x2=128y is

A
16
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B
32
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C
8
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D
24
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Solution

## The correct option is D −24Tangent of slope m to the parabola y2=16x is y=mx+4m …(1) Equation (1) is also a tangent to x2=128y, so x2=128(mx+4m) should have equal roots in x i.e., for mx2−128m2x−512=0 D=0⇒(128)2m4+4×512m=0⇒128m(128m3+16)=0⇒128m3+16=0 (∵m≠0)⇒m=−12 Therefore, the equation of common tangent is y=−x2−8⇒x−16+y−8=1 Hence, the sum of intercepts =−16−8=−24

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