Question

Find the equation of tangent to the curve x² + y² = 5, where the tangent is parallel to the line 2x – y + 1 = 0

Answer 1

Equation of the curve is x² + y² = 5   ......(i)

Equation of the line is 2x – y + 1 = 0  ......(ii)

Differentiating (i) w.r.t. x, we get

`2x + 2y ("d"y)/("d"x)` = 0

∴ `("d"y)/("d"x) = (-2x)/(2y) = -x/y`

Let (x₁, y₁) be any point on the curve.

∴ Slope of the tangent at (x₁, y₁)

= `(("d"y)/("d"x))_((x_1,  y_1)`

= `-(x_1)/(y_1)`

Slope of the line 2x – y + 1 = `(-2)/(-1)` = 2

Tangent at (x₁, y₁) is parallel to the line (ii) and hence their slopes are equal.

∴ `-(x_1)/(y_1)` = 2

∴ x₁ + 2y₁ = 0    ......(iii)

From (i), we get

y² = 5 – x²

∴ y = `sqrt(5 - x^2)`

∴ `("d"y)/("d"x) = 1/(2sqrt(5 - x^2)) (0 - 2x)`

= `(-x)/sqrt(5 - x^2)`

∴ `(("d"y)/("d"x))_((x_1,  y_1)` = `(-x_1)/sqrt(5 - (x_1)^2)`

= slope of the tangent at (x₁, y₁)

∴ `(-x_1)/sqrt(5 - (x_1)^2` = 2

Squaring on both sides, we get

∴ `(x_1)^2/(5 - (x_1)^2` = 4

∴ (x₁)² = 20 – 4(x₁)²

∴ 5(x₁)² = 20

∴ (x₁)² = 4

∴ x₁ = ± 2

From (iii), when x₁ = 2, then 2 + 2y₁ = 0

∴ y₁ = – 1

and when x₁ = – 2, then – 2 + 2y₁ = 0

∴ y₁ = 1

Thus, (x₁, y₁) = (2, – 1) or (– 2, 1)

Now, equations of the tangents are

y + 1 = 2(x – 2) and y – 1 = 2(x + 2)

i.e., 2x –  y –  5 = 0 and 2x –  y + 5 = 0