Application of Derivatives 1

Equation of the line through the point (12,2) and tangent to the parabola y = -x22 + 2 and secant to the curve y = √4 - x2 is

(A) 2x + 2y - 5 = 0
(B) 2x + 2y - 3 = 0
(C) y-2 = 0
(D) None of these

4 Answers

865
Soumyadeep Basu ·

I think it will be (A).

145
Ranadeep Roy ·

2x +4y -9=0

1357
Manish Shankar ·

y-2=m(x-1/2) is the equation of the line

This line will satisfy the give parabola

2+mx-m/2=-x2/2+2

=> x2+2mx-m=0

b2-4ac=0 will give
4m2+4m=0
m=0, m=-1

145
Ranadeep Roy ·

modifying the eqn of parabola
-2(y-2)=x2
so vertex(0,2)
therefore a=-1/2
general form of eqn of tangent to a parabola
(x-h)=m(y-k) + a/m
here h=0, y=2, a=-1/2
also
the tangent passes through x=1/2, y=2
so we get 1/2=-1/2m
so m=-1
therefore y-2x-1/2=-1
we get 2x +2y -5=0

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