Differentiate y2+2x−5=0 implicitly:
2y⋅dxdy+2=0⇒dxdy=−y1.
At (h,k), slope =−1/k.
Slope of line x+2y=4 is −1/2.
For parallel: −1/k=−1/2⇒k=2.
The tangent to the curve y2+2x−5=0 at the point (h, k) is parallel to the line x+2y=4, then the value of 'k' is:
Held on 4 Aug 2022 · Verified 13 Jul 2026.
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