Calculus

Application of Derivatives

Easy

The tangent to the curve $y^2 + 2x - 5 = 0$ at the point (h, k) is parallel to the line $x + 2y = 4$ , then the value of 'k' is:

$-\frac{1}{2}$

-2

$\frac{1}{2}$

✅ Correct Option: 3

Differentiate $y^2 + 2x - 5 = 0$ implicitly:

$2y \cdot \frac{dy}{dx} + 2 = 0 \Rightarrow \frac{dy}{dx} = -\frac{1}{y}$ .

At $(h, k)$ , slope $= -1/k$ .

Slope of line $x + 2y = 4$ is $-1/2$ .

For parallel: $-1/k = -1/2 \Rightarrow k = 2$ .

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The tangent to the curve y2+2x−5=0y^2 + 2x - 5 = 0y2+2x−5=0 at the point (h, k) is parallel to the line x+2y=4x + 2y = 4x+2y=4, then the value of 'k' is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The tangent to the curve $y^2 + 2x - 5 = 0$ at the point (h, k) is parallel to the line $x + 2y = 4$ , then the value of 'k' is: