865I think it will be (A).
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
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4 Answers
1452x +4y -9=0
Sourup Nag can u please provide the solution?Upvote·0· Reply ·2013-06-23 01:38:11- Ranadeep Roy Is d correct? But I didnt find any use of that secant part.
- Ranadeep Roy differentiating the eqn of parabola, we get dy/dx=slope=-x
Now the required tangent passes through x=1/2, so we get the slope=-1/2
So, single point form eqn of a straight line will give
(y-2)/(x-1/2)=-1/2
- Ranadeep Roy differentiating the eqn of parabola, we get dy/dx=slope=-x
Now the required tangent passes through x=1/2, so we get the slope=-1/2
So, single point form eqn of a straight line will give
(y-2)/(x-1/2)=-1/2
- Rudra Sen Do we really need the secant part?
- Sourup Nag A is the correct answer, but can someone please provide a solution as well?
- Soumyadeep Basu @Ranadeep, dy/dx=slope=-x. This x is not the x-coordinate of (1/2, 2) but the x-coordinate of the point through which the tangent passes. We do need the secant part as we can draw two tangents from any external point of a parabola.
1357y-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
145modifying 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