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

⇒ 2x^{2 – 1} + 2y^{2 – 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)