Makapuʻu

Makapuʻu is the extreme eastern end of the Island of Oʻahu in the Hawaiian Islands, comprising the remnant of a ridge that rises 647 feet (197 m) above the sea. The cliff at Makapuʻu Point forms the eastern tip and is the site of a prominent lighthouse. The place name of this area, meaning "bulging eye" in Hawaiian, is thought to derive from the name of an image said to have been located in a cave here called Keanaokeakuapōloli. The entire area is quite scenic and a panoramic view is presented at the lookout on Kalanianaole Highway (State Rte. 72) where the roadway surmounts the cliff just before turning south towards leeward Oʻahu and Honolulu.

The Makapuʻu area is reached approximately 2 kilometers (1.2 mi) east of Waimānalo Beach on Kalanianaole Highway (State Rte. 72) or from the Honolulu side (south shore; Hawaiʻi Kai) travelling east along the same highway beyond Sandy Beach. The Makapuʻu Point State Wayside Park, a 38-acre (15 ha) roadside park, is about midway up the draw on the right-hand side coming from Hawaiʻi Kai.

Features of special interest in this area include:

Makapuʻu Beach and Waimānalo Bay

Makapuʻu Point and lighthouse from Makapuʻu Beach

View of the Ka ʻIwi wilderness area from trail to Makapuʻu Point summit

"Rabbit" Island: the emerging lop-eared rabbit is swimming to the right

The islands across Makapuʻu

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Wikipedia