A tangent is drawn to the curve x2 + 2x –4ky + 3 = 0 at a point whose abscissa is 3.The tangent is perpendicular to the line x –2y + 3 = 0. Find the area bounded by the curve, this tangent, the x- axis and the ordinate x = –1.
Text Solution
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Sol. We have,
x 2 + 2x –4ky + 3 = 0
Differentiating with respect to x, we get
2x + 2 –4k 
= 0
⇒ 
= 
⇒ 
= 
= 
So, the slope of the tangent to the curve
x 2 + 2x –4ky + 3 = 0 at x = 3 is 
It is given that the tangent at x = 3 is perpendicular to the line x –2y + 3 = 0. Therefore,
× 
= –1 ⇒ k = –1
Putting k = –1 in x 2 + 2x –4ky + 3 = 0, we obtain
x 2 + 2x + 4y + 3 = 0
⇒ (x + 1) 2 = – 4 
which is a parabola with its vertices at (–1, –1/2) and opens downward.
Putting x = 3 in the equation of the parabola x 2 + 2x + 4y + 3 = 0 we get y= – 
The equation of the tangent at (3, – 
) is
y + = –2 (x –3) or, 2x + y –
= 0
In figure, we have to find the area of the shaded region.
Required area = Area of the region ABCD + Area of the region AED
= 
+ =
+
= 
+
=
sq. units.
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