Calculus 10 TH edition Ex 2.8 . Complete stepwise solutions from Question No 1 to Question No 47 with diagram to explain the geometrical properties. The main topic is:
DIFFERENTIATING EQUATIONS TO RELATE RATES
A rotated ellipse C whose equation is x2 − xy + y 2 = 4 . Show that the line y = x intersects C at
two points P and Q and that the tangent lines to C at P and Q are parallel.
Solution
x2 − xy + y 2 = 4 ............ eq(1)
Putting y=x in equation (1) =⇒
x2 − x2 + x2 = 4 OR x2 = 4 =⇒ x = 2, x = −2
Putting values of x =2 in equation (1)
22 − 2y + y 2 = 4 =⇒ 4 − 2y + y 2 = 4 =⇒ −2y + y 2 = 4 − 4 =⇒ y 2 − 2y = 0
=⇒ y(y − 2) = 0 =⇒ y = 0 Or y = 2.
Similarly putting x = −2 in equation (1)
(−2)2 + 2y + y 2 = 4 =⇒ 4 + 2y + y 2 = 4 =⇒ y 2 + 2y = 0
y(y + 2) = 0 −→ y = 0 Or y = −2
Since x = y =⇒ Only possible values are
(2, 2) and (−2, −2)
. Let (2, 2) be the point P and (−2, −2) be the point Q where curve C intersect the line Now
differentiating eq(1)
d(x2 −xy+y 2 ) d(4)
dx = dx =0
2 2
d(x −xy+y )
dx =0
dx2 d(xy) dy 2 dy dx dy 2 dy
dx − dx + dx = 2x − (x dx + y dx )+ dy dx =0
dy dy dy dy
2x − x dx − y + 2y dx = 0 =⇒ −x dx + 2y dx = y − 2x
dy dy y−2x
(2y − x) dx = y − 2x =⇒ dx = 2y−x .......... eq(2)
Slope of tangent line to the curve C at point P.That is at (x, y) = (2, 2)
dy 2−4
dx = 4−2 = −1
Slope of tangent line to the curve C at point Q.That is at (x, y) = (−2, −2)
1
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