Two straight lines are perpendicular to each other one of them touches the parabola y2=4a(x+a) and the other touches y2=4b(x+b). The locus of the point of intersection of these two lines is
A
x+a+b=0
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B
x−a−b=0
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C
x=0
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D
x+a=0
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Solution
The correct option is Ax+a+b=0 Let m be the slope of one line, then slope of other line will be −1m Equation of the line touching the parabola y2=4a(x+a) is y=m(x+a)+am⋯(1) Equation of the line touching the parabola y2=4b(x+b) is y=−1m(x+b)+b−1m y=−1m(x+b)−bm⋯(2) Solving equation (1) and (2) we get am+m(x+a)=−bm−1m(x+b) ⇒x(m+1m)+a(m+1m)+b(m+1m)=0 ∴x+a+b=0[∵m+1m≠0]