11mentor bhaiya, int the answer to q2 the intercept should be 12 and -27/4
Q1. Slope of common tangent to parabolas y2 = 4ax and x2= 8y
Q2.if 2y = x + 24 is atangent to the parabola y2 = 24x then its distance from parallel normal is
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4 Answers
1357Q2.if 2y = x + 24 is atangent to the parabola y2 = 24x then its distance from parallel normal is
y=mx-2am-am3 is the equation of normal to parabola y2=4ax
here a=6
eqn. of normal is y=mx-12m-6m3
slope of normal is (1/2) from the eqn of tangent
eqn of normal is y=x/2-6-3/4 or 4y=2x-27
distance=[24-(27/2)]/√5=21/2√5
1357Q1. Slope of common tangent to parabolas y2 = 4ax and x2= 8y
y=mx+a/m, x=my+2/m or mx=m2y+2
so m2=1 and 2=-a/m
so m=+1 then a=-2
m=-1 then a=2
slope is +1 if a=-2
and -1 if a=2
111@above
while finding the distance, coeff of x and y shld be same,this is what he has done
it seems to be right?? where is the prob??