Find the point on the curve x2 + y2 = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.
Given:
The curve x2 + y2 = 13 and the line 2x + 3y = 7
x2 + y2 = 13
Differentiating the above w.r.t x
⇒ 2x2 – 1 + 2y2 – 1
= 0
⇒ 2x + 2y
= 0
⇒ 2(x + y
) = 0
⇒ (x + y
) = 0
⇒ y
= – x
⇒ 
...(1)
Since, line is 2x + 3y = 7
⇒ 3y = – 2x + 7
⇒ y = 
⇒ y = 
+ 
The equation of a straight line is y = mx + c, where m is the The Slope of the line.
Thus, the The Slope of the line is 
...(2)
Since, tangent is parallel to the line,
the The Slope of the tangent = The Slope of the normal
= 
⇒ – x = 
⇒ x = 
Substituting x = 
in x2 + y2 = 13,
⇒ (
)2 + y2 = 13
⇒ (
) + y2 = 13
⇒ y2(
) = 13
⇒ y2(
) = 13
⇒ y2(
) = 1
⇒ y2 = 9
⇒ y = 
3
Substituting y = 
3 in x = 
,we get,
x = 
x = 
2
Thus, the required point is (2, 3) & ( – 2, – 3)